NCERT Exemplar: Nuclei- 1

# 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 = mproton + melectron
(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 = MP + Me - (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
(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. Mx and My 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 me denotes the mass of an electron, then which of the following statements is correct?
(a) Q1 = (Mx – My) c2 and Q2 = (Mx – My – 2me)c2
(b) Q1 = (Mx – My) c2 and Q2 = (Mx – My )c2
(c) Q1 = (Mx – My – 2me) c2 and Q2 = (Mx – My +2 me)c2
(d) Q1 = (Mx – My + 2me) c2 and Q2 = (Mx – My +2 me)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- My)c2
β+ decay is represented as:

⇒ Q2 = (M- M-2me)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. λA and λB 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 λA = λB
(b) Initial rate of decay of A is twice the initial rate of decay of B and λA > λB
(c) Initial rate of decay of B is twice the initial rate of decay of A and λA > λB
(d) Initial rate of decay of B is same as the rate of decay of A at t = 2h and λB < λA
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 λA > λB.
Also, when initial rate of decay of B is the same as rate of decay of A at t = 2h and λB < λA.

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> λB.  at point P the intersecting point of two graphs at time t same.

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

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, NA = N0 (maximum) whole NB = 0. As time increases, NA decreases exponentially and the number of atoms of B increases. After some time NB 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.
T1/2 of C-14 nuclei=5760 years
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 1.24×10eV 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
Z= N= 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

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=N0e1t
It is the rate of formation of Cl38 from S38. Let N2 is the number of of Cl38 atoms (radioactive):

1N0e1t2N2    ...(I)
Multiplying both sides by e2t 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 E2t to both sides we get

Nare the number of S38 atoms
No. Are Cl38 atoms after time ti will be N2=N0e2t

Put the value of N2 in (III)

By cross multiplication and multiplying both sides by

Number Cl38 radioactive atoms will be maximum at N= 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,
∴ Pn + pp= momentum of γ-ray of Energy E

Case I: If E=B then from

It can be possible if pn= pp=0 because square of non zero number can never zero.
If pn=pp=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),

For a real and equal value of Pp discriminant must be zero as the value of pp 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=En=mncand momentum of neutron pn=0 as its velocity is zero.
By the law of conservation of momentum,
Pn=pp+ p(Beta)
O= pp+ pe
Let pe=pp then
⇒ |pp|=|pe|=p(eV)
Energy of proton = Ep
Energy of electron
From conservation,

∴ By the law of conservation of energy,

and mnc2 =938 MeV
and mec2= 0.5 meV
As the energy difference in neutron and proton is very small, pc will be small pc<<mpcwhile pc may be greater than mec2 so by neglecting (mec2)2= (0.5)(given)

Again pc<<mpc2 so neglecting  we get
Pc = mnc2-mpc=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
t1/2 = 0.6931/λ = 0.6931/1.05
t1/2 =  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
Sp = (MZ–1, N + MH – MZ,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) Sp = (MZ–1, N + MH – MZ,N )c2
Here in this formula MZ-1 is the mass of atom of Z-1 atomic number.
Mis the mass of atom of mass number Z.
∴ MZ-1= Mass of atom whose atomic number is 50-1=49.
It is 49In119 in this case MZ-1 = 49In119 = 118.9058 and N=119 - 49 = 70.
Sp for 50Sn120=c2[118.9058+1.0078252-199.902199]
Sp for 50Sn120=0.0114362c2
Now for Sp of 51Sb121
Sp= [Mz-1, N+MH-Mz,N]c2
Z=51, Z-1=50 for Sn
MZ-1=mass of 50Sn=199.902199 u
∴ Sp for 51Sb121= [199.902199+1.0078252-120.903824]c2
= 0.0059912c2
∴ Sp(50Sn120)>Sp(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.

The document NCERT Exemplar: Nuclei- 1 | Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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## Physics Class 12

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## FAQs on NCERT Exemplar: Nuclei- 1 - Physics Class 12 - NEET

 1. What is a nucleus in the context of NCERT Exemplar: Nuclei?
Ans. The nucleus refers to the central core of an atom that contains protons and neutrons. It is responsible for the stability and mass of the atom.
 2. How does the number of protons in the nucleus affect the element's identity?
Ans. The number of protons in the nucleus determines the element's identity. It is known as the atomic number and represents the unique characteristic of each element.
 3. What is nuclear stability, and what factors contribute to it?
Ans. Nuclear stability refers to the balance between the strong nuclear force and the electrostatic repulsion between protons in the nucleus. Factors such as the number of neutrons and protons play a crucial role in determining nuclear stability.
 4. How does radioactive decay occur in the nucleus?
Ans. Radioactive decay occurs when an unstable nucleus undergoes spontaneous changes to become more stable. It releases energy in the form of radiation during this process.
 5. How does nuclear fission differ from nuclear fusion?
Ans. Nuclear fission involves the splitting of a heavy nucleus into two lighter nuclei, releasing a large amount of energy. On the other hand, nuclear fusion is the process of combining two light nuclei to form a heavier nucleus, also releasing a significant amount of energy.

## Physics Class 12

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