NCERT Exemplar Class 12 Physics Solutions Chapter 13 Nuclei

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

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 -radioactivity but not for others
(b) change for -radioactivity but not for others
(c) do not change for any type of radioactivity
(d) change for α and radioactivity but not for γ-radioactivity

Answer:

The answer is the option (b) change for and 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 = (M_x – M_y)c² and Q₂ = (M_x – M_y )c²
(b) Q₁ = (M_x – M_y)c² and Q₂ = (M_x – M_y – 2me )c²
(c) Q₁ = (M_x – M_y– 2me)c² and Q₂ = (M_x – M_y +2me )c²
(d) Q₁ = (M_x – M_y + 2me )c² and Q₂ = (M_x – M_y +2me )c²

Answer:

The answer is the option (b) Q₁ = (M_x – M_y) c² and Q₂ = (M_x – M_y – 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. A and B 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 A > B
(c) Initial rate of decay of A is twice the initial rate of decay of B and A = B
(d) Initial rate of decay of B is twice the initial rate of decay of A and A > B

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

and nuclei have the same mass number. Do they have the same binding energy?

Answer:

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

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 = -N
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, A.NA = B.NB
or, Na > NB
or, B > A
We know, = 1/
So,
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

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^-t
t = 2391.20 year

Question:19

Are the nucleons fundamental particles, or do they consist of still smaller parts? One way to find out is to probe a nucleon just as Rutherford probed an atom. What should be the kinetic energy of an electron for it to be able to probe a nucleon? Assume the diameter of a nucleon to be approximately 10^–15 m.

Answer:

We know that, = h/p and kinetic energy = potential energy
E = hc/
Kinetic energy of an electron
KE = PE = hc/ = 10⁹ eV

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

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 = . At what value of t, would the number of counts be a maximum?

Answer:

If we consider and as the disintegration constants for S38 and Cl38 respectively.
Therefore, dN₁/dt = – N₂
dN₂/dt = rate of decay of Cl38 + rate of formation of Cl38
again, dN₂/dt = N₂+ N1
edN₂ + N₂edt = N₀^{e(λ₂ – )}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 K_n, K_p, respectively.
pn and pp are the momenta of neutron and proton
E – B = K_n + K_p
B must be B²/4mc² so that B will be much bigger than E.

Question:23

The deuteron is bound by nuclear forces just as the H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in a deuteron as given in the form of a Coulomb potential but with an effective charge . Estimate the value of (e’/e) given that the binding energy of a deuteron is 2.2 MeV.

Answer:

H atom has a binding energy of = E = 13.6 eV
The value of m’, when it is reduced = 918 m
m_p = the mass of a neutron or a proton
The deuteron has a binding energy of 2.2 MeV
Therefore, the e’/e = 3.64

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 -decay. The neutron was at rest before -decay. Therefore, energy of neutron E_n = m_nc^₂ and momentum(p_n) = 0
After -decay, momentum of neutron (p)_n= momentum of proton (p_p) + momentum of electron (p_e)
pc= m_nc² – m_pc² = 938 MeV – 936 MeV = 2 MeV
Therefore, Energy of proton, E_p = 936 MeV
Therefore, energy of electron, E_e = 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:

(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 = –
Therefore, the value of = -(-4.16-3.11)/1 = 1.05 h-1
Now, Half-time(T1/2) = 0.693/ = 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 S_p = (M_Z-1, N + M_H – M_Z,N)c². Given In l_n¹¹⁹ = 118.9058u, S_n1²⁰= 119.902199u, S_b¹²¹ = 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:
S_p.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, S_p.S_n > 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.