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These solutions are designed by our experienced subject teachers as per the latest NCERT Textbook syllabus based om the CBSE Curriculum. Ace up your preparation with these NCERT Exercise questions and solutions PDF for class 12 physics and improve your subject knowledge on Chapter Semiconductor Electronics Materials Devices And Simple Circuits.

Class 12 Physics NCERT Solutions Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits

Students of 12th class can learn all the topics and subtopics of Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits from the NCERT Solutions. Semiconductor Electronics Materials Devices And Simple Circuits Concept speak about the types of semiconductors and insulators.

Moreover, it also includes various other important topics like conductors, classification of metals, and semiconductors. Discussion of the difference between the intrinsic semiconductor and extrinsic semiconductor, semiconductor diode, application of junction diode, etc will be seen in this chapter 14 NCERT Questions & Solutions.

Refer to the 12th Class Physics NCERT Solutions of Ch 14 Solved Exercise and Miscellaneous Exercise Questions at your preparation & revision time to clear all your doubts on Chapter Semiconductor Electronics Materials Devices And Simple Circuits.

Class	12
Subject	Physics
Book	Physics
Chapter Number	14
Chapter Name	Semiconductor Electronics Materials Devices And Simple Circuits

NCERT 12th Physics Ch 14 Solved & Miscellaneous Exercises Questions with Solutions PDF

Students who are preparing for their board exams or any other competitive exams can refer to these Class 12th Physics NCERT Solutions of Ch 14 Semiconductor Electronics Materials Devices And Simple Circuits. These solutions are taken from the NCERT Textbooks and our subject experts prepared these NCERT Solutions in a comprehensive way.

Question 1.
In an n-type silicon, which of the following statement is true :
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the doplants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) ‘Holes are minority carriers and pentavalent atoms are the dopants in n type semiconductor.’

Question 2.
Which of the statements given in Exercise 1 is true for p-type semiconductors ?
Answer:
(d) Holes are majority carriers and trivalent atoms are the dopants in p-type semiconductors.

Question 3.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c, (Eg)si and (Eg)Ge-
Which of the following statements is true ?
(a) (Eg)Si < (Eg)Ge < (Eg)c
(b)(E)c<(Eg)Ge>(Eg)si
(c) (Eg)c > (Eg)si > (Eg)Ge
(d) (Eg)c = (Eg)si = (Eg)Ge
Answer:
(C) (Eg)c > (Eg)Si > (Eg)Ge. Energy band gap is maximum in carbon and least in germanium among the given elements.

Question 4.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Answer:
(c) hole concentration in p-region is more as compared to n-region because hole diffusion takes place from higher concentration to lower concentration.

Question 5.
When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier
(b) reduces the majority carrier current to zero
(c) lowers the potential barrier
(d) none of the above.
Answer:
(c) lowers the potential barrier by cancelling the depletion layer.

Question 6.
For transistor action, which of the following statements are correct :
(a) Base, emitter and collector regions should have similar size and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse biased.
(d) Both the emitter junction as well as the collector junction are forward biased.
Answer:
(b) and (c) : The base region must be very thin, lightly doped and the emitter junction is forward biased whereas collector junction is reverse biased to avoid unnecessary diffusion of charge carrier in the base and also for proper amplification.

Question 7.
For a transistor amplifier, the voltage gain :
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
Answer:
(c) is low at high and low frequencies and constant at mid frequencies as per frequency response of a transistor.

Question 8.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Answer:
A half wave rectifier rectifies only one half cycle of input A.C.
.’. frequency of the output A.C.
= frequency of input A.C. = 50 Hz A full wave rectifier rectifies both halve cycles
of the A.C. input
.’. frequency of output A.C. = 2 x frequency of input A.C. = 2 x 50 = 100 Hz

Question 9.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kQ is 2 V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Answer:


Question 10.
Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is
0.01 volt, calculate the output ac signal.
Answer:


Question 11.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm ?
Answer:

Since the energy of the light photon is less than the band gap energy of p-n diode, it can not be detected.

Question 12.
The number of silicon atoms per m³ is 5 x 10²⁸. This is doped simultaneously with 5 x 10²² atoms per m³ of Arsenic and 5 x 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_i = 1.5 x 10¹⁶ m^-3. Is the material n-type or p- type ?
Answer:


Question 13.
In an intrinsic semiconductor the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K ? Assume that the temperature dependence of intrinsic carrier concentration n_i is given by

Answer:

Thus, conductivity of a semiconductor increases with rise in temperature.

Question 14.
In a P-n junction diode, the current I can be expressed as

where I₀ is reverse saturation current. V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, K_B is the Boltzmann constant (8.6 x 10^-5 eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 x 10^-12 A and T = 300 K, then
(a) What will be the forward current at forward voltage of 0.6 V ?
(b) What will be the increase in current if voltage across diode is increased to 0.7 V ?
(c) What is the dynamic resistance ?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V ?
Answer:
Statement of the given question is incorrect. The relation should be

Question 15.
You are given the two circuits as shown in Figure. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Answer:


Question 16.
Write the truth table for a NAND gate connected as given in Fig. Hence identify the exact logic operation carried out by these circuits.

Answer:
The NAND gate shown in the truth table has only one input. Therefore, the truth table is

A	A	y = A.A¯
0	0	1
1	1	0

Since Y = A¯ in this case, the circuit is actually a NOT gate with the truth table

A	Y
0	1
1	0

Question 17.
You are given two circuits as shown in Fig., which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:

Question 18.
Write the truth table for circuit given in the Fig. below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

Answer:
Let y₁ be the output which appears at the first operation of NOR gate.

A	B	Y
0	0	0
1	0	1
0	1	1
1	1	1

Question 19.
Write the truth table for the circuits given in the Fig., consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits

Answer:

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NCERT Solutions for Class 12 Physics Chapter 13 Nuclei PDF is one of the smartest study resources for all students out there during their preparation. It not only helps CBSE students but also state boards students like UP, Bihar, HP, MP, Gujarat, and many others. It supports you all in understanding the concepts clearly as they are written in a simple language.

So, we as a team from Ncertbook.guru advice you to refer this article and download NCERT Questions and solutions of ch 13 Nuclei for free in PDF format. You can access them anywhere and everywhere you wish during the preparation and get a great command on the physics Nuclei concept.

Class 12 Physics NCERT Solutions Chapter 13 Nuclei

In this chapter, you will deal with the important concept which is the core of atoms ie., Nuclei. From the Nuclei introduction to various other topics are discussed efficiently in the NCERT Solutions for class 12 physics chapter 13 Nuclei. Students will gain complete knowledge about the topics and subtopics of Nuclei from the important questions listed over in the NCERT Solutions PDF.

Class	12
Subject	Physics
Book	Physics
Chapter Number	13
Chapter Name	Nuclei

NCERT Exercise Questions & Answers of 12th Class Physics Chapter 13 – Nuclei

12th Class Physics NCERT Solutions of Chapter 13 Nuclei available over here and they are prepared by subject experts as per the CBSE Board Guidelines. You will find detailed answers to the class 12 physics textbook questions along with ch 13 nuclei exemplary problems, worksheets, and exercises here which aids you to grasp the complete knowledge on the topics of Nuclei.

Question 1.
(a) Two stable isotopes of lithium 63Li and 73Li have respective abundance of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u respectively. Find the atomic weight of lithium.
(b) Boron has two stable isotopes 105Li and 115Li ^.Their respective masses are 10.01294 u and 11.00931 u and the atomic weight of boron is 10.811 u. Find the abundances of 105Li and
115Li
Answer:
(a) Atomic weight of lithium


Question 2.
The three stable isotopes of neon :1020Ne and 1022Ne  have respective abundance of 90.51%, 0.27% and 9.22%. The atomic masses of three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.
Answer:

Question 3.
Obtain the binding energy of a nitrogen nucleus (147N) from the following data :
m_H = 1.00783 u
m_n = 1.00867 u
_mn = 14.00307 u
Give your answer in MeV.
Answer:

Question 4.
Obtain the binding energy of the nuclei 5626Fe and
in units of 20983Bi from the following data:
m_H =1007825u
m_n =1008665u
m (5626Fe)= 55.934939 u
m (20983Bi)
Which nucleus has greater binding energy per nucleon ?
Answer:

Question 5.
A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 6329Cu atoms (of mass 62.92960 u). The masses of proton and neutron are 1.00783 u and 1.00867 u, respectively.
Answer:

Question 6.
Write nuclear equations for :
(a) the α-decay of 22686Ra
(b) the β^–-decay of 3215p
(c) the β⁺-decay of 116p
Answer:

Question 7.
A radioactive isotope has a half-life of T years. After how much time is its activity reduced to 3.125% of its original activity (b) 1% of original value ?
Answer:

Question 8.
The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive 146C present with the stable carbon isotope
612C When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of 146C , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 146C dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.
Answer:

Question 9.
Obtain the amount of 6027Co necessary to provide a radioactive source of 8.0 mCi strength. The half­ life of 6027Co is 5.3 years.
Answer:

Question 10.
The half life of 9038Sr is 28 years. What is the disintegration rate of 15 mg of this isotope ?
Answer:

Question 11.
Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au and silver isotope 10747Au.
Answer:

Question 12.

Answer:

Question 13.

Answer:

Here m_N stands for the nuclear mass of the element or particle. In order to express the Q value in terms of the atomic masses, 6 m_e mass has to be subtracted
from atomic mass of 116Au and 5 m_e mass has to be

Question 14.
The nucleus 2310Ne decays by β~ emission. Write down the p-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m(2310Sr) = 22.994466 u
m(2311Sr) = 22.989770 u.
Answer:

Question 15.
The Q value of a nuclear reaction A + b -> C + d is defined by
[Q = m_A + m_b-m_c– m_d] c² where the masses refer to nuclear rest masses. Determine from the given data whether the following reactions are exothermic or endothermic.

Answer:

Question 16.
Suppose, we think of fission of a 5626Fe nucleus into two equal fragments, if 2813Al. Is the fission energetically possible ? Argue by working out Q of the process. Given, m (5626Fe) = 55.93494 u and m (2813Al)= 27.98191
Answer:


Question 17.
The fission properties of 23994Pu are very similar to those of 23592uu. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure
23994Pu undergo fission ?
Answer:

Question 18.
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much
23592u did it contain initially ? Assume that all the energy generated arises from the fission of 23592u and that this nuclide is consumed by the fission process.
Answer:

Question 19.
How long an electric lamp of 100 W can be kept glowing by fusion of 2.0 kg of deuterium ? The fusion reaction can be taken as

Answer:

Question 20.
Calculate the height of Coulomb barrier for the head on collision of two deuterons. The effective radius of deuteron can be taken to be 2.0 fm.
Answer:
The initial mechanical energy E of the two deutrons before collision is given by
E = 2 K.E.

Question 21.
From the relation R = R₀ A^1/3, where R₀  is a constant and A is the mass number of a nucleus, show that nuclear matter density is nearly constant (i.e. independent of A)
Answer:

Question 22.
For the β⁺ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K- shell, is captured by the nucleus and a neutrino is emitted).

Show that if β⁺ emission is energetically allowed, electron capture is necessarily allowed but not vice-versa
Answer:

Question 23.
In a Periodic Table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on Earth. The three isotopes and their masses are 2412Mg (23.98504u), ? 2512Mg (24.98584) and 2612Mg (25.98259u). The natural abundance of 2412Mg is 78.99% by mass. Calculate the abundances of the other two isotopes.
Answer:

Question 24.
The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 2412Ca and 2713Al from the following data :

Answer:

Question 25.

Answer:

Question 26.
Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes :

(a) Calculate the Q values for these decays and determine that both are energetically possible.
(b) The Coulomb barrier height for α-particle

Answer:

Question 27.
Consider the fission of 23992u by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products, after the beta-decay of the primary fragments, are 14058Ce and 9944Ru. Calculate Q for this fission process. The relevant atomic and particle masses 

Answer:

Question 28.
Consider the D-T reaction (deuterium-tritium-fusion) given in eqn. :

(b) Consider the radius of both deuterium and tritium to be approximately 1.5 fm. What is the kinetic energy needed to overcome the Coulomb repulsion ? To what temperature must the gases be heated to inititate the reaction ?
Answer:
From the equation given in the question,

m_N refer to nuclear mass of the element given in the brackets and m_n = mass of neutron. If in represents the atomic mass, then


Question 29.
Obtain the maximum kinetic energy of p-particles, and the radiation frequencies to y decays in the following decay scheme. You are given that
m (¹⁹⁸Au) = 197.968233 u
m (¹⁹⁸Hg) = 197.966760 u
Answer:
The total energy released for the transformation of
19879Au to 19880u can be found by considering the energies of ϒ-rays. We first find the frequencies of the ϒ-rays emitted.




Question 30.
Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun and (b) the fission of 1.0 kg of ²³⁵U in a fission reactor.
Answer:

Question 31.
Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plant. Suppose we are given that, on an average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium did our country need per year by 2000 ? Take the heat energy per fission of ²³⁵U to be about 200 MeV. Avogadro’s number = 6.023 x 10²³ mol^-1.
Answer:

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Candidates who are searching for NCERT solutions for Class 12 Physics Chapter 12 – Atoms can refer to this page completely. Here, we as a team covered all topics and subtopics solved questions of atoms and presented in these 12th class physics NCERT Solutions Chapter 12 Atoms PDF for a better understanding of the concept.

In NCERT Solutions, each and every question comes with a detailed, step-by-step solution in easy and simple language for all concepts and subjects that helps you to improve your skills. So, go ahead and download the NCERT Solutions for Class 12 Physics Chapter 12 Atoms in pdf and prepare well for all examinations.

Class 12 Physics NCERT Solutions Chapter 12 Atoms

In Chapter 12, Students will acquire the basic knowledge about Atoms in detail. An Atom is the smallest component of an element, portrayed by a sharing of the chemical properties of the element and a nucleus with neutrons, protons, and electrons. You will have a complete grip on atom chapter concepts like Rutherford’s atomic model, atomic spectra, Bohr’s model of the hydrogen atom, and the line spectra in the hydrogen atom, etc with NCERT Solutions.

Ace up your exam preparation with this Class 12 Physics Exercisewise NCERT Solutions for Chapter 12 Atoms and score max. marks in both CBSE Board exams and Competitive exams.

Class	12
Subject	Physics
Book	Physics
Chapter Number	12
Chapter Name	Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms – Solved Exercises

Studying from the NCERT Solutions for Class 12 Physics Ch 12 is the best way to cover all the topics included in the Atoms concept. Having a thorough practice helps you to score well in the Physics CBSE Board Exam paper. Also, you can understand & get familiar with all the topics and subtopics important questions of chapter Atoms preparing from NCERT Solutions Pdf.

Question 1.
Choose the correct alternative from clues given at end of the each statement:
(a) The size of the atom in Thomson’s model is the atomic size in Rutherford’s model.(much greater than/no different from/much less than.)
(b) In the ground state of…………… electrons are in stable equilibrium, while in ……………..  electrons always experience a net force. (Thomson’s model/Rutherford’s model.)
(c) A classical atom based on…………. is doomed to collapse.
(Thomson’s model/Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a………….. but has a highly non­uniform mass distribution in  (Thomson’s model/Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in………….
(Rutherford’s model/both the models.)
Answer:
(a) no different from
(b) Thomson’s model; Rutherford’s model.
(c) Rutherford’s model.
(d) Thomson’s model, Rutherford’s model.
(e) both the models.

Question 2.
Suppose you are given a chance to repeat the alpha- particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperature 14 K). What results do you expect ?
Answer:
α-particle is much heavier than hydrogen nuclei. Therefore, α-particles pass through solid hydrogen without deflection. In other words, α-particles are not scattered by solid hydrogen.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines ?
Answer:
The wavelength of the spectral lines forming Paschen series is given by

Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper level to the lower level ?
Answer:

Question 5.
The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state ?
Answer:
K.E. = -E (Total energy)
= -(-13.6) = 13.6 eV
P.E. = 2 X E = 2 X (-13.6)
= -27.2 eV

Question 6.
A hydrogen atom initially in the ground level absorbs a photon which excites it to the
n = 4 level. Determine the wavelength and frequency of photon.
Answer:
We know, energy of an electron in the nth orbit of hydrogen atom is given by

Question 7.
Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the
n = 1,2 and 3 levels. (b) Calculate the orbital period in each of these levels.
Answer:
(a) Speed of an electron in nth orbit of hydrogen atom is given by

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 x I0^-11 m. What are the radii of the n = 2 and n = 3 orbits ?
Answer:
We know, radius of nth orbit of hydrogen atom is given by
r_n = r₀n², where r₀ = 5.3 x 10^-u m is the radius of
the innermost orbit of hydrogen atom.
When n = 2, r₂ = 5.3 x 10^-u X 4
= 2.12 X 10-¹⁰m
When n = 3,
= 5.3 x 10-¹¹ x 9
= 4.77 x 10-¹⁰m.

Question 9.
A 12.75 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted ?
Answer:


Question 10.
In accordance with the Bohr’s model, And the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 x 10¹¹ m with orbital speed
3 x 10¹⁴ m s^-1. (Mass of earth = 6.0 x 10²⁴ kg.)
Answer:
According to Bohr’s postulate of quantization of angular momentum

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model ?
(b) Is the probability of backward scattering (i.e., scattering of a-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model ?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of a-particles scattered at moderate angles is proportional to What clue does this linear independence on t provide ?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a-particles by a thin foil ?
Ans:
(a) It is about the same because we are considering average deflection angle.
(b) It is much less than the predicted value because the massive core is absent in Thomson’s model.
(c) It suggests that the scattering of a-particles is primarily because of the single collision.
(d) In Thomson’s model of atom. Multiple collisions are required to be considered in this model because positive charge is spread throughout in this model.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10^-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
If electron and proton were bound by gravitational attraction, then

It is astonishing this value of r is much greater than the size of universe.

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom
de-excites from level it to level (n – 1). For large it, show that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The energy of an electron in the nth orbit of hydrogen atom is given by

i.e. frequencies are equal. This is called the Bohr’s correspondence principle.

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size ? Why is an atom not, say, thousand times bigger than its typical size ? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To stimulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10-¹⁰m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, m_e, and c. Determine its numerical value.
(b) You will find that the length obtained in («) many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, m_e, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, m_g, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Here, the dimensional formula of e is A¹T¹, the dimensional formula of m_e is M¹, dimensional formula

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about
– 3.4 eV.
(a) What is the kinetic energy of the electron in this state ?
(b) What is the potential energy of the electron in this state ?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed ?
Answer:
(a) K.E. = -E = – (-3.4 eV) = 3.4 eV
E. = 2E = 2 x (-3.4 eV)
= -6.8 eV
(b) Kinetic energy does not depend upon the choice of zero of potential energy. Therefore, its value remains unchanged. However, the potential energy gets changed with the change in the zero level of potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun ?
Answer:
Applying the Bohr’s quantization postulate,

i.e., n is very large. Since n is very large, the difference between the two successive energy or angular momentum levels is very small and the levels may be considered continuous.

Question 17.
Obtain the first Bohr’s radius and the ground state energy of a ‘muonic hydrogen atom’ (i.e. an atom in which a negatively charged muon (μ^-1) of mass about 207 m_e orbits around a proton).
Answer:
Here mass of the particle revolving around the proton is

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NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter Free PDF is provided on this page. All these solutions help students of class 12 in attempting numerical problems and theory questions on the board and entrance exams. You can find all Dual Nature of Radiation and Matter important questions solutions in NCERT Solutions of 12th Physics Ch 11 material.

Subject experts have given all these solutions with a straightforward and detailed description. So that you can rely on these Class 12 Physics NCERT Solutions for Chapter 11 PDF at your exam preparation. Dual Nature of Radiation and Matter NCERT Solutions PDF covers the complete syllabus in a smart manner and solutions are prepared as per the latest CBSE guidelines.

Class 12 Physics NCERT Solutions Chapter 11 Dual Nature of Radiation and Matter

Class 12 Physics Chapter 11 NCERT Solutions covers 9 topics and two main arguments of the Dual Nature of Radiation and Matter. This concept discusses some of the main topics and subtopics like Electron Emission, Photoelectric Effect, Hertz’s observations, Hallwachs’ and Lenard’s observations, the wave theory of light, wave nature of matter, etc.

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Class	12
Subject	Physics
Book	Physics
Chapter Number	11
Chapter Name	Dual Nature of Radiation and Matter

NCERT Questions with Solutions for Class 12 Physics Chapter 11

Class 12 Physics NCERT Solutions Ch 11 Dual Nature of Radiation and Matter plays a vital role in scoring max marks in the physics paper. So, candidates of different boards are advised to follow & prepare from the NCERT Solutions for Class 12 physics Ch 11. Because NCERT Solutions are prepared by the subject experts based on the CBSE board prescribed NCERT Textbooks.

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴ Hz is incident on the metal surface, photo emission of electrons occurs. What is the
(а)   maximum kinetic energy of the emitted electrons,
(b)   stopping potential, and
(c) maximum speed of the emitted photo electrons ?
Answer:

Question 3.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?
Answer:
Here, V₀ = 1.5 V
Maximum kinetic energy. (K.E.)_max
= eV₀ = 1.6 X 10-¹⁹ X 1.5
= 2.4 x 10-¹⁹ J

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on the average, arrive at a target irradiated by this beam ? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon ?
Answer:
Given, λ = 632.8 nm = 632.8 x 10^-9m
Power, P = 9.42 mW = 9.42 x 10^-3 W


Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 x 10³ W/m². How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:

Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10^-15 V s. Calculate the value of Planck’s constant.
Answer:

Question 7.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with the sodium light ?
(b) At what rate are the photons delivered to the sphere ?
Answer:
Power of sodium lamp,
P = 100 W

Question 8.
The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If light of frequency
8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm ?
Answer:

Since energy of incident radiation less than the work function (4.2 eV) of the metal, therefore, the photoelectric emission can not take place from the given metal. 11.10.

Question 10.
Light of frequency 7.21 x 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 10⁵ m/s are ejected from the surface. What is the threshold frequency for photo emission of electrons ?
Answer:
Using the relation

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photo electrons is 0.38 V. Find the work function of the material from which the cathode is made.
Answer:

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Given V = 56 V.
(a) momentum of the electron

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV ?
Answer:
Here, E = 120 eV = 120 x 1.6 x 10^-19J
= 1.92 x 10-¹⁷ J

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.
Answer:

Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 x 10^-9 kg drifting with a speed of 2.2 m/s ?
Answer:
(a) Here, m = 0.040 kg and υ = 1.0 km/s .
= 1000 m/s

Question 16.
An electron and a photon each have a wavelength of 100 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.
Answer:

Question 17.
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 x 10-¹⁰ m?  (C.B.S.E. 2008)
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \(\frac { 3 }{ 2 } \) kT at 300 K.
Answer:
(a) Here λ = 1.40 x 10^-10m
Also, h = 6.63 X 10^-34 Js
and m = 1.67 x 10^-27kg



Question 18.
Show that the wavelength of electromagnetic radiaion is equal to the de Broglie wavelength of its quantum (photon).
Answer:

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer:
For nitrogen,


Question 20.
(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The ‘specific charge’ of the electron i.e., its elm is given to be 1.76 x 10¹¹ C kg^-1.
(b) Use the same formula you employ in
(a) to obtain electron speed for an anode potential of 10 MV. Do you see what is
wrong ? In what way is the formula to be modified ?
Answer:




Question 21.
(a) A monoenergetic electron beam with electron speed of 5.20 x 10⁶ ms^-1 is subject to a magnetic field of l.30 x 10^-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given elm for electron equals 1.76 X 10¹¹ C kg^-1.
(b) Is the formula you employ in
(a) valid for calculating radius of the path of a 20 MeV electron beam ? If not, in what way is it modified ?
Answer:
(a) υ  = 5.20 x 10⁶ms^_1
B = 1.30 x 10 ⁴T

(b) The formula employed in part (a) is not valid because with the increase in velocity, mass varies and in the above formula we have taken m as constant. Instead, m

Question 22.
An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10^-2 mm of Hg). A magnetic field of
2.83 X 10 ⁴ T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture ; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:

Question 23.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at (45 Å. What is the maximum energy of a photon in the radiation ?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube.
Answer:

Question 24.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron- positron pair of total energy 10.2 BeV into two y-rays of equal energy. What is the wavelength associated with each γ-ray ?   (1 BeV = 10⁹ eV)
Answer:
Energy carried by the pair of γ-rays = 10 .2 BeV
Energy of each γ ray is

Question 25.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons ! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^_1° W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 x 10¹⁴
Answer:


This is quite a small number, but still large enough to be counted.
Comparison of cases (a) and (b) tells us that our eye can not count the number of photons individually.

Question 26.
Ultraviolet light of wavelength 2271 A from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity
(~ 10^s W m^-2) red light of wavelength 6328 A produced by a He-Ne laser ?
[C.B.S.E. 2005, UC, 13]
Answer:

Since the energy of red photon is less than the work function for the metal, photo cell does not respond to red light.

Question 27.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon lamp irradiates photosensitive material made of caesium or tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:

Question 28.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
λ_x = 3650 Å, λ₂ = 4047 Å, λ₃ = 4358 Å, λ₄ = 5461 Å, λ₅ = 6907 Å.
The stopping voltages, respectively, were measured to be :
V₀₁ = 1.28 V, V₀₂ = 0.95 V, f₀₃ = 0.74 V, = 0.16 V,V₀₅ = 0 V
(a) Determine the value of Planck’s constant
(b) Estimate the threshold frequency and work function for the material.
Answer:
(a) From the Einstein photoelectric equation,




Question 29.
The work function for the following metals is given : Na : 2.75 eV ; K : 2.30 eV ; Mo : 417 eV ; Ni: 515 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away ? (C.B.S.E. 2009)
Answer:
Here, λ = 3300 A = 3300 x 10^-10 m
Distance, r’ = 1 m and r’ = 50 cm = 0.5 m
Using the relation

Since the energy E of the incident photon of light is less than the work functions of Mo and Ni metals, so photoelectric emission will not occur in Mo and Ni. The distance of the source does not increase or decrease the energy of the photon of light incident, therefore, the energy of electrons ejected will not change but the intensity of ejected electrons will increase (1 α 1/r² ) and become four times.

Question 30.
Light of intensity 10^-5 W m^-2 falls on a sodium photo­cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave- picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer ?
Answer:
A = 2 cm² = 2 x 10^-4 m²
φ = 2eV = 2 x 1.6 x 10^-19 J
= 3.2 X 10^-19  J.
Taking the approximate radius of an atom as 10^-10 m. the effective area of sodium atom is ≈ r² = 10^-20 m.
.’. If there is one free electron per atom, then number of electrons in five layers


The answer obtained implies that the time of emission of electron is very large and is not in agreement with the observed time of emission, which is approximately 10 ⁹ s. Thus wave-picture of radiation is not applicable for photo-electric emission.

Question 31.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy ? (For quantitative comparison, take the wavelength of the probe equal to 1 Å which is of the order of inter-atomic spacing in the lattice) (m_e = 9.11 x 10^­-31 kg).
Answer:

Clearly, the energy of photon is much greater than the energy of electron.

Question 32.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 44, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ?
Explain. (m_n = 1.675 x 10^-27 kg).
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer:

This wavelength is about hundred times smaller than the interatomic separation of crystals. Thus, neutrons are not suitable for diffraction experiments in case of crystals.

This wavelength is comparable to the interatomic spacing of crystals. Therefore, thermal electrons are able to interact with the crystal. Since

increasing the temperature, decreases their de Broglie wavelength and they become unsuitable for crystal diffraction. Thus, the fast beam of neutrons needs to be thermalised with the environment for neutron diffraction experiment.

Question 33.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow
light ?
Answer:



Question 34.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^-15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV).
Answer:

Thus, the energy of the proton ejected out of the linear accelerator is of the order of BeV.

Question 35.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure ; and compare it with the mean separation between two atoms under these conditions.
Answer:
Here T = 27 °C = 27 + 273 = 300 K
P = 1 atm = 1.01 x 10⁵ Nm^-2
Also, mass of helium atom,


Question 36.
Compute the typical de Broglie wavelength of an electron in a metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10^_10 m.
Answer:
Here, T = 27°C = 27 + 273 = 300 K


Question 37.
Answer the following questions :
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e; (- l/3)e]. Why do they not show up in *Millikan’s oil-drop experiment?
(b) What is so special about the combination elm ? Why do we not simply talk of e and m separately ?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures ?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic ? Why is there an energy distribution of photoelectrons ?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :

Answer:
(a) In case of the Millikan oil drop experiment, the charge on the electron is measured. The electron revolve outside the nucleus and each has a charge e. Thus, we do not observe the fractional charges

(c) At ordinary pressure, molecules of gas keep on colliding with each other and the ions formed do not have a chance to reach the respective electrodes to constitute a current because of their recombination. At low pressure, however, ions do not collide frequently and are able to reach the respective electrodes to constitute a current.
(d) Work function in fact is the energy required to knock out the electron from highest filled level of conduction band of an emitter. In the conduction band, there are different energy levels which collectively form a continuous band of levels. Therefore, different amounts of energy are required to bring the electrons out of the different levels. Electrons emitted have different kinetic energies according to the energy supplied to the emitter.
(e) Since frequency for a given matter wave remains constant for different layers of the matter but wavelength changes so X is more significant than v.
Similarly energy E = hv = \(\frac { 1 }{ 2 } \) m(λv)² is also constant for a given matter wave so phase λv is also not physically significant.

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You will avail the answers in a detailed explanation of each concept included in Ch 4 Moving charge and Magnetism class 12 physics NCERT Solutions. These solutions are prepared by subject experts based on the latest ncert books for physics and CBSE guidelines. Going through with these NCERT Solved Questions of 12th Physics Ch 4 helps you get strong basics on the concept of Moving charge & Magnetism and it leads to score well in the exams.

Class 12 Physics NCERT Solutions Chapter 4 Moving Charge and Magnetism

Chapter 4 is so important because many of the coming learnings will have their base. So, pay some more attention while practicing this ch 4 Moving charge and magnetism concept. This Class 12 Physics NCERT chapter 4 illustrates what a magnetic force is and it’s motion in a magnetic field in a detailed manner. A total of 11 topics are addressed in this Physics Class 12 Moving charge and Magnetism Chapter.

Mostly it talks about ampere’s circuital law and Biot-savarat’s law. Also, it proposes students to the toroid and the solenoid, motion in combined magnetic and electric fields, etc. So, Class 12 Physics NCERT Solutions for Chapter 4 Moving Charges and Magnetism presented here can aid students to clarify all their queries instantly.

Class	12
Subject	Physics
Book	Physics
Chapter Number	4
Chapter Name	Moving Charge and Magnetism

NCERT Exercise Questions & Answers of 12th Class Physics Ch 4

12th Class NCERT Solutions for Physics chapter 4 by Ncertbooks.guru is a reliable study material for your learning process. Class 12 Physics NCERT Solutions of Chapter 4 Moving Charges and Magnetism benefits CBSE class 12 Science students to an extent & make them feel confident while attempting the final exam.

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the center of the coil ?
Answer:

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the Held B at a point 20 cm from the wire ?
Answer:

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of R at a point 2.5 m east at the wire.
Answer:

Question 4.
A horizontal overhead power line carries a current of 90A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line ?
Answer:

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T ?
Answer:

Question 6.
A 3.0 cm wire carrying a current of 10A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire ?
Answer:
Using F = BIl, we get
F = 0.27 x 10 x 3 x 10 x 10^-2
= 8.1 x 10-² N
The direction of force is perpendicular to the direction of magnetic field as well as to the direction of current.

Question 7.
Two long and parallel straight wires A and B carrying currents of 8*0 A and 5*0 A in the same direction are separated by a distance of 4-0 cm. Estimate the force on a 10 cm section of wire A.
Answer:

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its center.
Answer:

Question 9.
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil ?
Answer:
Using τ= NIAB sin θ, we get
τ = 20 x 12 x 10 x 10 x 10^-4 x 0.8 x sin 30
= 0.96 Nm

Question 10.
Two moving .coil meters, M₁and M₂ have the following particulars:
R₁ = 10 Ω, N₁ = 30, A₁= 3.6 x 10^-3m², B₁= 0.25 T
R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 x 10^-3 m², B₂ = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Answer:
Using Current sensitivity = NBA/k
For M₁ Current Sensitivity


Question 11.
In a chamber, a uniform magnetic field of 6.5G (1G = RHT) is maintained. An electron is shot into the field with a speed of 4.8 x 10⁶ ms^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit,
(e =1.6 x 10^-19 C, m = 9.1 x 10^-31 kg).
Answer:

Question 12.
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron ? Explain.
Answer:

Question 13.
(a) A circular coil of 30 turns and radius 8-0 cm carrying a current of 6-0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar toil of some irregular shape that encloses the same area ? (All other particulars are also unaltered)  (C.B.S.E. 1998 C )
Answer:
(a) Using τ = NBIA sin 9, we get
τ= 30 X 1 x 6 x n (8 X 10^-2)² sin 60
= 180 x it (8 x 10-²)² 0.866
= 3.13 N m
The magnitude of the counter torque is 3 .13 N m
(b) Answer will not change because torque does not depend upon shape of the coil provided it encloses same area.

Question 14.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A : coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their center.
Answer:
For Coil X

Question 15.
A magnetic field of 100 G (1 G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^-3 m². The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^-1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Answer:

We may take I = 10 A and n = 800. The given solenoid may have length of 50 cm having 400 turns and area of cross-section = 5 x 10^-3m² (five times the given value.)

Question 16.
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its center is given by,

(a) Show that this reduces to the familiar result for field at the center of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

(Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.)
Answer:

(b) Let there be two coils as mentioned in the statement. Magnetic field in a small region of length 2d about the mid-point of the space between the two coils is given by,


Question 17.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (i) outside the toroid, (ii) inside the core of the toroid, and (iii) in the empty space surrounded by the toroid.
Answer:
(i) Zero

Question 18.
Answer the following questions :
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle ?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment ?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflection from its straight line path.
Ans.
(a) The particle goes undeflected so it does not experience magnetic force.
Using F = qυ B sin θ. It is clear that the initial velocity u is either parallel or anti-parallel to magnetic field (B) i.e. θ is either 0° or 180°.

(b) Yes. the final speed is equal to the initial speed because the magnetic force can change the direction of o but the magnitude of o can not be changed by it.

(c) Force acting on electron due to electrostatic field ( E→ ),
F→_e =-eE→

The direction of this force is towards North. Therefore, electron is deflected towards North. Force acting on electron due to magnetic field, F→_m= -e v→ xB→. The electron will not be deflected and move in a straight line path, if the force due to magnetic field on the electron is towards South. According to Fleming’s left hand rule, the direction of B→ is along the vertical in downward direction.

Question 19.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Answer:
K.E, acquired by electron while passing through V

(b) When electron moves with velocity r making an angle of 30° with the direction of magnetic field, then r cos θ is


Question 20.
A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10⁵ V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique.
Answer:

The given particle may be deutron.
The result is not unique because this e/m ratio can be true for He^{+ +} , Li ^{+ + +} etc.

Question 21.
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.”
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero ?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before ? (Ignore the mass of the wires.) g = 9.8 ms^-2.
Answer:
(a) The tension in the wire is zero if force on the current  carrying wire due to current is equal and opposite to the weight of the wire. This is, BIl= mg

(b) In case current is reversed, the tension is equal to the force acting on the wire due to magnetic field plus weight of the wire. This is,
T = BlL + mg
= 0.26 x 5 x 0.45 + 60 x 10^-3 x 9.8
= 1.18 N.

Question 22.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart ? Is the force attractive or repulsive (H.S.E.B.2001)
Answer:

Since the currents in two wires are in opposite direction so the force is repulsive.

Question 23.
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction.
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm ?
Answer:

Question 24.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Figure? What is the force on each case ? Which case corresponds to stable equilibrium ?

Answer:

Question 25.
A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field ?
(The coil is made of copper wire of cross-sectional area 10^-5 m², and the free electron density in copper is given to be about 10²⁹ m³)
Answer:

Question 26.
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its center) normal to its
axis ; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? g = 9.8 ms^-2.
Answer:

Question 27.
A galvanometer coil has a resistance of 12 Q and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V ?
Answer:
Here 1=3 mA = 3 x 10^-3 A
Galvanometer resistance, G = 12 Ω The galvanometer can be converted into the voltmeter of range 0 to V (here V = 18 V) by connecting a high series resistance R given

Question 28.
A galvanometer coil has a resistance of 15Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 ?Answer:

In this chapter 4 certain topics have an allotment of good marks like magnetic field and direction of a circular coil, magnetic force, etc. So, give some extra time to practice all these important concepts questions and answers covered in the NCERT Textbooks for class 12 Physics. You can 100% rely on the NCERT solutions for 12th class physics ch 4 Moving charge and Magnetism during the exam prep and score max. marks in the physics board exam paper.

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