NCERT Exemplar: Nuclei- 1 | Physics Class 12 - NEET PDF Download

MULTIPLE CHOICE TYPE QUESTIONS I

Q.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 have 5000 atoms of the material. 
(b) All the containers will contain the same number of atoms of the material but that number will only be approximately 5000. 
(c) The containers will in general  have different numbers of the atoms of the material but their average will be close to 5000. 
(d) None of the containers can have more than 5000 atoms.
Ans. (c)
Solution.
Half life time for a radioactive substance is defined as the time in which a radioactive atomic substance remains half of its original value of radioactive atom. So after one year means one half life i.e., average atoms of radioactive substance remain after 1 Year in each container is equal to 1/2 of 10,000 = 5000 atoms (average).

Q.2. The gravitational force between a H-atom and another particle of mass m will be given by Newton’s law:
 where r is in km and
(a) M = m_proton + m_electron
(b)
(c) M is not related to the mass of the hydrogen atom.
(d)  (|V| = magnitude of the potential energy of electron in the H-atom).
Ans. (b)
Solution.
During formation of H-atom some mass of nucleons convert into energy by E = mc², this energy is used to bind the nucleons along with nucleus. So mass of atom becomes slightly less than sum of actual masses of nucleons and electrons.
Actual mass of H atom = M_P + M_e - (B.E./c²) (B/c² is binding energy)
B.E. (B) of H atoms is 13.6 eV per atom.

Q.3. When a nucleus in an atom undergoes a radioactive  decay, the electronic energy levels of the atom
(a) Do not change for any type of radioactivity
(b) Change for α and β radioactivity but not for γ-radioactivity
(c) Change for α-radioactivity but not for others
(d) Change for β-radioactivity but not for others
Ans. (b)
Solution.
β- Particles carries one unit of negative charge, and α-particle carries 2 units of positive charge, and Υ-particle carries no charge. So the electronic energy level of the atom changes in emission of α and β particle, but not in Υ decay.

Q.4. M_x and M_y denote the atomic masses of the parent and the daughter nuclei respectively in a radioactive decay. The Q-value for a β^– decay is Q₁ and that for a β⁺ decay is Q₂. If m_e 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 – 2m_e)c²
(b) Q₁ = (M_x – M_y) c² and Q₂ = (M_x – M_y )c²
(c) Q₁ = (M_x – M_y – 2m_e) c² and Q₂ = (M_x – M_y +2 m_e)c²
(d) Q₁ = (M_x – M_y + 2m_e) c² and Q₂ = (M_x – M_y +2 m_e)c²
Ans. (a)
Solution.
Key concept: Q value or energy of nuclear reaction: The energy absorbed or released during a nuclear reaction is known as Q-value of nuclear reaction.
Q-value = (Mass of reactants – mass of products)c² Joules
= (Mass of reactants – mass of products) amu
If Q < 0, the nuclear reaction is known as endothermic. (The energy is absorbed in the reaction)
If Q > 0, the nuclear reaction is known as exothermic. (The energy is released in the reaction)
Let the nucleus be _ZX⁴.
β^- decay is represented as:

⇒ Q₁ = (M_x - M_y)c²
β⁺ decay is represented as:

⇒ Q₂ = (M_x - M_y -2m_e)c²

Q.5. Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into . If one of the neutrons in Triton decays, it would transform into He³ nucleus. This does not happen. This is because
(a) Triton energy is less than that of a He³ 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) Because free neutrons decay due to external perturbations which is absent in a triton nucleus
Ans. (a)
Solution.
Triton (₁H³) has 1 proton and 2 neutrons. If a neutron decays as: , then nucleus will have 2 proton and 1 neutron, i.e. triton atom converts in ₂He³ (2 proton and 1 neutron).
Binding energy of ₁H³ is much smaller than ₂He³  so transformation is not possible energetically.

Q.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
Ans. (b)
Solution.
Key concept: Neutron-proton ratio (N/2 ratio:) The chemical properties of atom are governed entirely by the number of protons (Z ) in the nucleus, stability of an atom appears to depend on both the number o f protons and number of neutrons.
(i) For lighter nuclei, the greatest stability is achieved when the number o f protons and neutrons are approximately equal (N = Z ), i.e. N/Z = 1
(ii) Heavy nuclei are stable only when they have more neutrons than protons. Thus heavy nuclei are neutron rich compared to lighter nuclei ( for heavy nuclei, more is the number of protons in the nucleus, greater is the electrical repulsive force between them. Therefore more neutrons are added to provide the strong attractive forces necessary to keep the nucleus stable.)

Q.7. In a nuclear reactor, moderators slow down the neutrons which come out in a fission process. The moderator used 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 liquid or gaseous state at room temperature
Ans. (b)
Solution. 
Key concept: A moderator is a material used in a nuclear reactor to slow down the neutrons produced from fission. By slowing the neutrons down the probability of a neutron interacting with Uranium-235 nuclei is greatly increased thereby maintaining the chain reaction. Moderators are made from materials with light nuclei which do not absorb the neutrons but rather slow them down by a series of collisions.
The moderator only slows neutrons down in order to increase the interaction with Uranium nuclei. They do not give any protection if the reaction goes out of control. 1 fa chain reaction is heading out of control the reactors needs to be able to reduce the concentration of neutrons. For this the reactor uses control rods. Control rods are matte from material with the ability to absorb neutrons. Cadmium and Boron are examples of suitable materials. By inserting.control rods between the fuel rods the chain reaction can be slowed dowp-or shut down. Withdrawing the control rods can restart or speed up the reaction.
In our given question, the moderator used have light nuclei (like proton). When protons undergo perfectly elastic collision with the neutron emitted their velocities are exchanged, i.e., neutrons come to rest and protons move with the velocity of neutrons.
Heavy nuclei will not serve the purpose because elastic collisions of neutrons with heavy nuclei will not slow them down.

MULTIPLE CHOICE TYPE QUESTIONS II

Q.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
Ans. (a, b)
Solution.
Two deuteron can combine to form He atom when their nuclei come close to nuclear range where electrostatic repulsive force between positively charged deuterons does not act. Electrostatic force increases very high on decreasing their distance

To overcome this electrostatic repulsive force nuclei need very high temperature and pressure. Hence to combine two nuclei, they must reach closer of the range of where nuclear force acts and electrostatic repulsive force does not act verifies the answer (a) and (b).

Q.9. Samples of two radioactive nuclides A and B are taken. λ_A and λ_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 A is twice the initial rate of decay of B and λ_A = λ_B
(b) Initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B
(c) Initial rate of decay of B is twice the initial rate of decay of A and λ_A > λ_B
(d) Initial rate of decay of B is same as the rate of decay of A at t = 2h and λ_B < λ_A
Ans. (b, d)
Solution.
Key concept:
Law of radioactive disintegration : According to Rutherford and Soddy law for radioactive decay is as follows:
“At any instant the rate of decay of radioactive atoms is proportional to the number of atoms present at that instant.” i.e.
dN/dt ∞ N ⇒ dN/dt = -λN
it can be proved that N=N₀e^-λ1          
In terms of mass M— M₀e^-λ1   
where N = Number of atoms remains undecayed after time t,
N₀ = Number of atoms present initially (i.e., at t = 0),
M = Mass of radioactive nuclei at time t,
M₀ = Mass of radioactive nuclei at time t = 0,
N₀-N= Number of disintegrated nucleus in time t,
dN/dt= rate of decay, λ = Decay constant or disintegration constant or radioactivity constant or Rutherford Soddy’s constant or the probability of decay per unit time of a nucleus.
The samples of the two radioactive nuclides A and B can simultaneously have the same decay rate at any time if initial rate of decay of A is twice the initial rate of decay of B and λ_A > λ_B.
Also, when initial rate of decay of B is the same as rate of decay of A at t = 2h and λ_B < λ_A. 

Q.10. The variation of decay rate of two radioactive samples A and B with time is shown in Figure.

Which of the following statements are true?
(a) Decay constant of A is greater than that of B, hence A always decays faster than B.
(b) Decay constant of B is greater than that of A but its decay rate is always smaller than that of A.
(c) Decay constant of A is greater than that of B but it does not always decay faster than B.
(d) 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.
Ans. (c, d)
Solution.
From the given graph slope of A is greater than B so rate of decay of A is greater than of  or instant t or for a particular time  so λ_A > λ_B.  at point P the intersecting point of two graphs at time t same.

VERY SHORT ANSWER TYPE QUESTIONS

Q.11. He₂³ and He₁³ nuclei have the same mass number. Do they have the same binding energy?
Ans. The nuclei He₂³ and He₁³ have the same mass number. He₂³ has two protons and one neutron. He₂³ has one proton and two neutrons. As He3 has only one proton hence the repulsive force between protons is missing in ₁He³, so the binding energy of ₁He³ is greater than that of ₂He³. 

Q.12. Draw a graph showing the variation of decay rate with number of active nuclei.
Ans.
According to Rutherford and Soddy law for radioactive decay  where decay constant (λ) is constant for a given radioactive material. Therefore, graph between N and dN/dt is a straight line as shown in the diagram.

Q.13. Which sample, A or B shown in Figure has shorter mean-life?
Ans. Initially at t=0 from figure given

i.e., initially both samples has equal number of radioactive atoms. Considering at any instant t = t from figure,



So mean life time of sample A is greater than of B.

Q.14. Which one of the following cannot emit radiation and why?  Excited nucleus, excited electron.
Ans. Excited electron cannot emit radiation because energy of electronic energy levels is in the range of eV and not MeV (mega electron volt), y-radiations have energy of the order of MeV.
Key concept: The energy of internal motion of a nucleus is quantized. A typical nucleus has a set of allowed energy levels, including a ground state (state of lowest energy) and several excited states. Because of the great strength of nuclear interactions, excitation energies of nuclei are typically of the order of the order of 1 MeV, compared with a few eV for atomic energy levels. In ordinary physical and chemical transformations the nucleus always remains in its ground state. When a nucleus is placed in an excited state, either by bombardment with high-energy particles or by a radioactive transformation, it can decay to the ground state by emission of one or more photons called gamma rays or gamma-ray photons, with typical energies of 10 keV to 5 MeV. This process is called gamma (γ) decay.

Q.15. In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved?
Ans.  when an electron and positron combine together coming from opposite directions they destroy each other by the emission of two Υ-rays in opposite direction to conserve linear momentum as below
 

SHORT ANSWER TYPE QUESTIONS

Q.16. Why do stable nuclei never have more protons than neutrons?
Ans. The reason is that protons, being charged particles, repel each other. This repulsion becomes so great in nuclei with more than 10 protons or so, that an excess of neutrons which produce only attractive forces, is required for stability.
Important point: As you get to heavier elements, with each new proton you add, there is a larger repulsive force. The nuclear force is attractive and stronger than the electrostatic force, but it has a finite range. So you need to add extra neutrons, which do not repel each other, to add extra attractive force. You eventually reach a point where the nucleus is just too big, and tends to decay via alpha decay or spontaneous fission.
To view this in quantum mechanical terms, the proton potential well is not as deep as the neutron well due to the electrostatic repulsion. [Due to the Pauli exclusion principle, you only get two particles per level (spin up and spin down)]. If one well is filled higher than the other, you tend to get a beta decay to even them out. As the nuclei get larger, the neutron well gels deeper as compared to the proton well and you get more neutrons than protons.

Q.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 number of atoms of A and B versus time.
Ans. Consider radioactive nucleus A have N₀ atoms of A initially; or at t = 0, N_A = N₀ (maximum) whole N_B = 0. As time increases, N_A decreases exponentially and the number of atoms of B increases. After some time N_B becomes maximum. As B is an intermediate nuclei which is also radioactive, it also start decaying and finally drop to zero exponentially by radioactive decay law. We can represent the situation as shown in the graph.

Q.18. A piece of wood from the ruins of an ancient building was found to have a ¹⁴C 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 half-life of ¹⁴C is 5760 years.
Ans. Rate of disintegration in old wood sample of C-14 radioactive atoms is 12 atoms per min gm. Initially rate of disintegration of C-14 when the tree was live=16 atoms per min per gm.
T_1/2 of C-14 nuclei=5760 years
According to radioactive decay law,
N=N₀e^-λt or R=R₀e^-λt
12=16e^-λt

Q.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.
Ans. To detect the properties of nucleons inside the nucleus the wavelength of particle which may detect nucleons that must be of size of nucleons (10^-15 m). So the wavelength of particle which can detect the nucleons must be equal to or less than 10^-15
λ = 10^-15 m
λ = h/p
∵ E = hv = hc/λ    [∵c = vλ]


=12.4×10^-34+42=1.24×10¹×10⁺⁸
K.E. = 1.24×10⁹eV
So the K.E. of particle which may detect nucleon inside the nucleus must be of 1.24×10⁹ eV per particle.

Q.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 ²³₁₁Na ?
(b) Which nuclide out of the two mirror isobars have greater binding energy and why?
Ans.
Here Z is atomic number and N is no. of neutron in ₁₁Na²³
Z₁=11
N₁=23-11=12
Mirror isobar of ₁₁Na²³ is
Z₂ = N₁ = 12
So Mg is isobar of ₁₁Na²³
So ₁₂Mg²³ is the mirror isobar of ₁₁Na²³
(b) As the neutrons in ₁₂Mg²³ are ‘11’ and in ₁₁Na²³ are ‘12’ so, the number of neutrons in Na is larger than Mg and hence nuclear short range attractive forces in Na will be larger than repulsive electrostatic forces between proton-proton.
So, ₁₁Na²³ has more binding energy than ₁₂Mg²³

LONG ANSWER TYPE QUESTIONS

Q.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?
Ans.

Initially at t=0, number of radioactive atoms of S³⁸ = N₁ and of Cl³⁸ are zero.
At any time t,
and N₁=N₀e^-λ1^t
It is the rate of formation of Cl³⁸ from S³⁸. Let N₂ is the number of of Cl³⁸ atoms (radioactive):

=λ₁N₀e^-λ1^t-λ₂N₂    ...(I)
Multiplying both sides by e^+λ2^t dt


Integrating both sides

∴ Cl³⁸ atom is formed after disintegration of S³⁸, so initially number of Cl³⁸ atoms are N₂=0.
at t=0, N₂=0,



Multiplying E^-λ2^t to both sides we get


N₀ are the number of S₃₈ atoms
No. Are Cl³⁸ atoms after time ti will be N₂=N₀e^-λ2^t


Put the value of N₂ in (III)

By cross multiplication and multiplying both sides by








Number Cl³⁸ radioactive atoms will be maximum at N₂ = 0.8267 hrs.

Q.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.
Ans. Binding energy (B) of deuteron = 2.2 MeV
Some part of energy of γ-ray is used up against binding energy B = 2.2 MeV and the rest part will impart K.E. to neutron and proton.

By law of conservation of momentum,
∴ P_n + p_p= momentum of γ-ray of Energy E

Case I: If E=B then from

It can be possible if p_n= p_p=0 because square of non zero number can never zero.
If p_n=p_p=0 then equation IInd cannot be satisfied and the process cannot take place. From II, but energy E of γ ray cannot be zero.
Case II: if E>B or E=B+λ where λ will be very small than B then from (I),




It is a quadratic equation so its solution by quadratic formula


For a real and equal value of P_p discriminant must be zero as the value of p_p must be one.



∵ λ is very small
So E=B


Q.23. The deuteron is bound by nuclear forces just as H-atom is made up of p and e bound by electrostatic forces. If we consider the force between neutron and proton in deuteron as given in the form of a Coulomb potential but with an effective charge e′  estimate the value of (e’/e) given that the binding energy of a deuteron is 2.2 MeV.
Ans. The binding energy of H atom in ground state

If proton and neutron had charge e’ each and governed by the same electrostatic force, then in the above equation we would need to replace electronic mass m by the reduced mass m’ of proton-neutron (as some mass of proton and neutron is used by binding energy) and electronic charge e is replaced by e’.


=M/2 ( if m = mass of electron)
m' = 1836m/2 = 918 m
∴ Binding Energy
Dividing (II) by (i) we get,

e'/e = 2200000/1248.48 = (176.21)^1/4
Required ratio e'/e = 3.64

Q.24. Before the neutrino hypothesis, the beta decay process was throught to be the transition,

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.
Ans.  Neutron was at rest before β decay from neutron. Hence energy of neutron=E_n=m_nc² and momentum of neutron p_n=0 as its velocity is zero.
By the law of conservation of momentum,
P_n=p_p+ p_e (Beta)
O= p_p+ p_e
Let p_e=p_p then
⇒ |p_p|=|p_e|=p(eV)
Energy of proton = E_p 
Energy of electron
From conservation,

∴ By the law of conservation of energy,

and m_nc² =938 MeV
and m_ec²= 0.5 meV
As the energy difference in neutron and proton is very small, pc will be small pc<<m_pc² while pc may be greater than m_ec² so by neglecting (m_ec²)²= (0.5)² (given)

Again pc<<m_pc² so neglecting  we get
Pc = m_nc²-m_pc² =938 MeV-936 MeV
Pc=2 MeV is the momentum
E=mc²
∵ E² = m²c⁴
E is the energy of either proton or neutron then


Q.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  versus t and obtain the value of half-life from the graph.
Ans.
(i) Graph between R versus t is exponential curve. From the graph at slightly more  the R should be 50% so at R =50% the  t(h)=0.7h
=0.7×60 min
=42 min

(ii) For Graph between  versus t(h)


=2.302 log₁₀ 0.3536=1.04
at t=2 hours,
= 2.303log₁₀ 0.125 = -2.08

t (hours)	1	2	3	4
	-1.04	-2.08	-311	-4.16

The graph showing the variation of  versus t(h) as follows:
We know that disintegration constant

λ = -1.05 per hour
t_1/2 = 0.6931/λ = 0.6931/1.05
t_1/2 =  42 min


Q.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 ¹²⁰Sn (Z = 50) and ¹²¹Sb = (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 ¹¹⁹In = 118.9058u, ¹²⁰Sn = 119.902199u, ¹²¹Sb = 120.903824u, ¹H = 1.0078252u.
(ii) What does the existence of magic number indicate?
Ans.
(i) S_p = (M_{Z–1, N} + M_H – M_Z,N )c²
Here in this formula M_Z-1 is the mass of atom of Z-1 atomic number.
M_Z is the mass of atom of mass number Z.
∴ M_Z-1= Mass of atom whose atomic number is 50-1=49.
It is ₄₉In¹¹⁹ in this case M_Z-1 = ₄₉In¹¹⁹ = 118.9058 and N=119 - 49 = 70.
S_p for ₅₀Sn¹²⁰=c²[118.9058+1.0078252-199.902199]
S_p for ₅₀Sn¹²⁰=0.0114362c²
Now for S_p of ₅₁Sb¹²¹
S_p= [M_z-1, _N+M_H-M_z,N]c²
Z=51, Z-1=50 for S_n
M_Z-1=mass of ₅₀Sn=199.902199 u
∴ S_p for ₅₁Sb¹²¹= [199.902199+1.0078252-120.903824]c²
= 0.0059912c²
∴ S_p(₅₀Sn¹²⁰)>S_p(₅₁Sb¹²¹)
(ii) The existence of magic numbers indicates that the shell structure of nucleus is similar to the shell structure of atom. This also explains the peaks in binding energy per nucleon curve.