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NCERT Exemplar Class 12 Physics Solutions Chapter 13 Nuclei

NCERT Exemplar Class 12 Physics Solutions Chapter 13 Nuclei

Edited By Safeer PP | Updated on Sep 14, 2022 01:28 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 13 discusses various properties of nuclei and defines it as a collection of positively charged particles called protons and neutral neutrons. Nucleus is a Latin word for a seed inside a fruit; it is the very dense central region of an atom. The nucleus's volume is 10^-12 times the atom's volume; the nucleus contains most of the atom's mass. NCERT Exemplar Class 12 Physics chapter 13 solutions provided on this page would help you prepare well for the 12 boards and competitive exams and clarify doubts regarding any concept based on nuclei. Students can go through NCERT and also make use of NCERT Exemplar Class 12 Physics solutions chapter 13 PDF download for further understanding of concepts.

Also see - NCERT Solutions for Class 12 Other Subjects

NCERT Exemplar Class 12 Physics Solutions Chapter 13 MCQI

Question:1

Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half-life of 1 year. After 1 year
(a) all the containers will contain the same number of atoms of the material but that number will only be approximately 5000
(b) all the containers will have 5000 atoms of the material
(c) none of the containers can have more than 5000 atoms
(d) the containers will, in general, have different numbers of the atoms of the material but their average will be close to 5000

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Answer:

The answer is the option (d) the containers will, in general, have different numbers of the atoms of the material but their average will be close to 5000.

Question:2

The gravitational force between an H-atom and another particle of mass m will be given by Newton’s law: F = G M.m/r2 , where r is in km and
(a) M = m proton + m electron
(b) M = m proton + m electron - (|V |/c2) (where, |V| = magnitude of the potential energy of electron in the H-atom)
(c) M is not related to the mass of the hydrogen atom
(d) M = m proton + m electron – (B/c2) (where, B = 13.6 eV)

Answer:

Answer: The answer is the option (d) M = m proton + m electron – B/c2 (where, B = 13.6 eV)

Question:3

When a nucleus in an atom undergoes radioactive decay, the electronic energy levels of the atom
(a) change for \alpha-radioactivity but not for others
(b) change for \beta-radioactivity but not for others
(c) do not change for any type of radioactivity
(d) change for α and \beta radioactivity but not for γ-radioactivity

Answer:

The answer is the option (b) change for \alpha and \beta radioactivity but not for γ-radioactivity

Question:4

Mx and My denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a - decay is Q1 and that for a + decay is Q2. If me denotes the mass of an electron, then which of the following statements is correct?
(a) Q1 = (Mx – My)c2 and Q2 = (Mx – My )c2
(b) Q1 = (Mx – My)c2 and Q2 = (Mx – My – 2me )c2
(c) Q1 = (Mx – My– 2me)c2 and Q2 = (Mx – My +2me )c2
(d) Q1 = (Mx – My + 2me )c2 and Q2 = (Mx – My +2me )c2

Answer:

The answer is the option (b) Q1 = (Mx – My) c2 and Q2 = (Mx – My – 2me )c2

Question:5

Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into P + ? + ? . If one of the neutrons in Triton decays, it would transform into He-3 nucleus. This does not happen. This is because
(a) because free neutrons decay due to external perturbations which is absent in a triton nucleus
(b) the electron created in the beta decay process cannot remain in the nucleus
(c) both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a He3 nucleus
(d) Triton energy is less than that of a He3 nucleus

Answer:

The answer is the option (d) Triton energy is less than that of a He3 nucleus

Question:6

Heavy stable nuclei have more neutrons than protons. This is because of the fact that
(a) neutrons are heavier than protons
(b) electrostatic force between protons are repulsive
(c) neutrons decay into protons through beta decay
(d) nuclear forces between neutrons are weaker than that between protons

Answer:

Answer: The answer is the option (b) electrostatic force between protons are repulsive

Question:7

In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used to have light nuclei. Heavy nuclei will not serve the purpose because
(a) they will break up
(b) elastic collision of neutrons with heavy nuclei will not slow them down
(c) the net weight of the reactor would be unbearably high
(d) substances with heavy nuclei do not occur in a liquid or gaseous state at room temperature

Answer:

The answer is the option (b) elastic collision of neutrons with heavy nuclei will not slow them down

NCERT exemplar class 12 physics solutions chapter 13 MCQII

Question:8

Fusion processes, like combining two deuterons to form a He nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact:
(a) nuclear forces have short-range
(b) nuclei are positively charged
(c) the original nuclei must be completely ionized before fusion can take place
(d) the original nuclei must first break up before combining with each other

Answer:

The correct answers are the options,
(a) nuclear forces have short-range
(b) nuclei are positively charged

Question:9

Samples of two radioactive nuclides A and B are taken. \lambdaA and \lambdaB are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaneously have the same decay rate at any time?
(a) Initial rate of decay of B is the same as the rate of decay of A at t = 2h and λB < λA
(b) Initial rate of decay of A is twice the initial rate of decay of B and \lambdaA > \lambdaB
(c) Initial rate of decay of A is twice the initial rate of decay of B and \lambdaA = \lambdaB
(d) Initial rate of decay of B is twice the initial rate of decay of A and \lambdaA > \lambdaB

Answer:

The correct answers are the options,
(a) Initial rate of decay of B is the same as the rate of decay of A at t = 2h and λB < λA
(b) Initial rate of decay of A is twice the initial rate of decay of B and λA > λB

Question:10

The variation of the decay rate of two radioactive samples A and B with time is shown in the figure. Which of the following statements are true?

(a) The decay constant of A is greater than that of B, hence A always decays faster than B
(b) The decay constant of B is greater than that of A but its decay rate is always smaller than that of A
(c) The decay constant of A is greater than that of B but it does not always decay faster than B
(d) The decay constant of B is smaller than that of A but still, its decay rate becomes equal to that of A at a later instant

Answer:

The correct answers are the options,
(c) The decay constant of A is greater than that of B but it does not always decay faster than B
(d) The decay constant of B is smaller than that of A but still, its decay rate becomes equal to that of A at a later instant

NCERT Exemplar Class 12 Physics Solutions Chapter 13 Very Short Answer

Question:11

He_{2}^{3} and He_{1}^{3} nuclei have the same mass number. Do they have the same binding energy?

Answer:

Both in He_{2}^{3} and He_{1}^{3}, there are the same mass numbers. However, the difference between these two nuclei is binding energy. In the case of He_{1}^{3}, the binding energy is more than that of He_{2}^{3}. This is because the number of protons and neutrons present in both of the nuclei are different. The He_{1}^{3} nuclei contain only one proton with two neutrons. However, the He_{2}^{3} nuclei have two protons and only one neutron in it.

Question:12

Draw a graph showing the variation of decay rate with the number of active nuclei.

Answer:


From the Rutherford and Soddy law, we get, the radioactive decay can be expressed as-\frac{dn}{dt}=\lambda N

Question:13

Which sample, A or B shown in the figure has shorter mean-life?

Answer:

From the above figure, we get:
At t= 0,
(dN0/dt)A = (dN0/dt)B
as, dN/dt = -\lambdaN
Therefore, (No)A = (No)B
Hence, at the beginning, there are the same numbers of radioactive atoms in both samples. Let, at any instant of time t,
(dN0/dt)A > (dN0/dt)B
or, \lambdaA.NA = \lambdaB.NB
or, Na > NB
or, \lambdaB > \lambdaA
We know, \tau = 1/\lambda
So, \tau_A>\tau_B
Here, sample B has a shorter mean life than sample A.

Question:14

Which one of the following cannot emit radiation and why? Excited nucleus, excited electron.

Answer:

From the above mentioned, an excited electron cannot emit radiation. In case of an excited electron, the energy level is in the eV range instead of MeV.

Question:15

In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved?

Answer:

An electron and a positron destroy each other to produce gamma radiation in case of pair annihilation. During this process, they used to go in opposite directions in order to conserve their momentum.

NCERT Exemplar Class 12 Physics Solutions Chapter 13 Short Answer

Question:16

Why do stable nuclei never have more protons than neutrons?

Answer:

In stable nuclei, the number of protons is lesser than the number of neutrons. This is because protons used to repel each other as they have a charge. Due to this high repulsion, the excess neutrons produce attractive forces. As a result, the stability is maintained.

Question:17

Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence: A → B → C Here B is an intermediate nuclei which is also radioactive. Considering that there are N0 atoms of A initially, plot the graph showing the variation of a number of atoms of A and B versus time.

Answer:


Here, in the above graph:
At t=0, NA = N0
NA falls off exponentially with the increase in time.
As a result, the number of atoms of B increases, it will be maximum once and then decays to zero at \infty

Question:18

A piece of wood from the ruins of an ancient building was found to have a C14 activity of 12 disintegrations per minute per gram of its carbon content. The 14C activity of the living wood is 16 disintegrations per minute per gram. How long ago did the tree, from which the wooden sample came, die? Given the half-life of C14 is 5760 years.

Answer:

As per the information given in the question, C14 activity of a piece of wood from the ruins is R = 12 dis/min per gram
C14 activity of a living wood = Ro = 16 dis/min per gram
The half-life of C14 = 5760 years
Using radioactive law,
R = Roe-\lambdat
t = 2391.20 year

Question:20

A nuclide 1 is said to be the mirror isobar of nuclide 2 if Z1 =N2 and Z2 =N1.
(a) What nuclide is a mirror isobar of 2311 Na?
(b) Which nuclide out of the two mirror isobars have greater binding energy and why?

Answer:

a) We can understand that a nuclide of 1 must be the mirror isobar of nuclide 2, when: Z1 = N2 and Z2 = N1
As a result, the mirror isobar is Z2 = 12 – N1
Similarly, N2 = 23 -12 = 11 = Z1
b) The binding energy of Mg is greater than Na.

NCERT Exemplar Class 12 Physics Solutions Chapter 13 Long Answer

Question:21

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is
Sulphur^{38\xrightarrow[=2.48h]{half-life}}Cl^{38}\xrightarrow[=0.62h]{half-life}Ar^{38}(stable)
Assume that we start with 1000 38S nuclei at time t = 0. The number of 38Cl is of count zero at t = 0 and will again be zero at t = \infty . At what value of t, would the number of counts be a maximum?

Answer:

If we consider \lambda_{1} and \lambda_{2} as the disintegration constants for S38 and Cl38 respectively.
Therefore, dN1/dt = – \lambdaN2
dN2/dt = rate of decay of Cl38 + rate of formation of Cl38
again, dN2/dt = \lambda_{2}N2+ \lambda_{1}N1
e\lambda_{2}tdN2 + \lambda_{2}N2e\lambda_{2}tdt = \lambda_{1}N0e(λ2\lambda_{1})dt
If we integrate the above equation, we will get:
t = 1.65 h

Question:22

Deuteron is a bound state of a neutron and a proton with a binding energy B = 2.2 MeV. A γ -ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident γ-ray. If E = B, show that this cannot happen. Hence calculate how much bigger than B must E be for such a process to happen.
Answer:

A deuteron has a binding energy B, = 2.2 MeV
The kinetic energies of neutrons and protons are Kn, Kp, respectively.
pn and pp are the momenta of neutron and proton
E – B = Kn + Kp
B must be B2/4mc2 so that B will be much bigger than E.

Question:24

Before the neutrino hypothesis, the beta decay process was thought to be the transition, N→ P+?. If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.

Answer:

Here we have to observe the before and after cases of the \beta-decay. The neutron was at rest before \beta-decay. Therefore, energy of neutron En = mnc2 and momentum(pn) = 0
After \beta-decay, momentum of neutron (p)n= momentum of proton (pp) + momentum of electron (pe)
pc= mnc2 – mpc2 = 938 MeV – 936 MeV = 2 MeV
Therefore, Energy of proton, Ep = 936 MeV
Therefore, energy of electron, Ee = 2.06 MeV

Question:25

The activity R of an unknown radioactive nuclide is measured at hourly intervals. The results found are tabulated as follows:

t(h)
0
1
2
3
4
R(MBq)
100
35.36
12.51
4.42
1.56
(i) Plot the graph of R versus t and calculate half-life from the graph.
(ii) Plot the graph of ln(R/R0) versus t and obtain the value of half-life from the graph.

Answer:


From the above graph, we can clearly understand that 50% reduction of R has happened. Therefore, the value of the half-life is 40 mins.
In the graph, the value of t = OB, that is equal to 40 mins.
ii)

The above graph is ln(R/R0) versus t.
Here, the slope of the graph = – \lambda
Therefore, the value of \lambda = -(-4.16-3.11)/1 = 1.05 h-1
Now, Half-time(T1/2) = 0.693/ \lambda = 0.66 h = 39.6 min

Question:26

Nuclei with magic no. of proton Z = 2, 8, 20, 28, 50, 52 and magic no. of neutrons N = 2, 8, 20, 28, 50, 82 and 126 are found to be very stable
(i) Verify this by calculating the proton separation energy Sp for Sn120(Z = 50) and Sb121 = (Z = 51). The proton separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by Sp = (MZ-1, N + MH – MZ,N)c2. Given In ln119 = 118.9058u, Sn120= 119.902199u, Sb121 = 120.903824u, H1 = 1.0078252u
(ii) What does the existence of a magic number indicate?

Answer:

i) Here, from the given data, the energy required for the separation of proton is:
Sp.Sn = (M119.70 + Mh – M120.70)c2 = 0.0114362 c2
Similarly, Sp.Sp = (M120.70 + Mh – M121.70)c2 = 0.0059912 c2
Here, we can clearly observe, Sp.Sn > Sp.Sp
Therefore, the Sn nucleus will be more stable than Sb nucleus.
ii) From the magic numbers, we can understand that the shell structure of the nucleus and the shell structure of the atom are similar. Peaks in the Binding energy can also be observed from it.

NCERT Exemplar Class 12 Physics Solutions Chapter 13 Nuclei-Main Subtopics

  • Introduction
  • Atomic Masses And Composition Of Nucleus
  • Size Of The Nucleus
  • Mass-energy And Nuclear Binding Energy
  • Mass – Energy
  • Nuclear binding energy
  • Nuclear Force
  • Radioactivity
  • Law of radioactive decay
  • Alpha decay
  • Beta-decay
  • Gamma decay
  • Nuclear Energy
  • Fission
  • Nuclear reactor
  • Nuclear fusion – energy generation in stars
  • Controlled thermonuclear fusion
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What will the students learn in NCERT Exemplar Class 12 Physics Solutions Chapter 13 Nuclei?

Class 12 Physics NCERT Exemplar solutions chapter 13 will look for answers to questions like - Does the nucleus have a structure? What are the constituents of the nucleus? How are these held together? We will study the behavioural and physical properties such as size, mass, and stability of the nuclei and the associated phenomena such as radioactivity, fission, and fusion. NCERT Exemplar solutions for Class 12 Physics chapter 13 also brushes through the concepts of isotopes, mass-energy, nuclear energy, nuclear force, and the different types of decay.

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Frequently Asked Question (FAQs)

1. What all topics are covered in this chapter?

 This chapter will cover all the topics related to nucleus, its properties, nuclear energy, its mass, and radioactivity.

2. How can these solutions be helpful?

  One can use these NCERT exemplar Class 12 Physics solutions chapter 13 to understand the chapter in more detail so that it can be easy to solve questions in the exam.

3. What all questions are solved in this chapter?

 We have solved all the main exercise questions and the additional questions given in the book. We have also solved both numerical and theoretical questions.

4. Who has solved these questions?

 We have highly experienced teachers who have major hold on the topic. They have designed these NCERT exemplar Class 12 Physics solutions chapter 13 nuclei as per student’s convenience.  

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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