Chapter 13: Nuclei

Q13.11Written answer

\(\begin{array}{l}He_{2}^{3}\end{array} \)

and
\(\begin{array}{l}He_{1}^{3}\end{array} \)
nuclei have the same mass number. Do they have the same binding energy?

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Worked solution

He2³ and He1³ have the same mass number but the binding energy of these two nuclei is different. The binding energy of the He1³ is greater than the He2³ because the number of protons and neutrons present in both the nuclei are different. He1³ has one proton and two neutrons while He2³ has two protons and one neutron.

Q13.12Written answer

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

dN/dt

N

According to Rutherford and Soddy law, the radioactive decay is given as –dN/dt = λN.

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

dN/dt

A

B

T

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Worked solution

At t= 0, (dN/dt)A = (dN/dt)B

dN/dt = -λN

(N_o)A = (N_o)B

λANA = λBNB

Na > NB

λB > λA

Q13.14Written answer

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

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Worked solution

An excited electron cannot emit radiation because the energy of the electronic energy level is in the range of eV and not in MeV.

Q13.15Written answer

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

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Worked solution

In pair annihilation, an electron and a positron destroy each other to produce gamma radiation and their momentum is conserved as they move in opposite directions to conserve the momentum.

Short Answers

Q13.16Written answer

Why do stable nuclei never have more protons than neutrons?

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A stable nuclei never have more protons than neutrons because protons are charged particles and they repel each other. The repulsion is so much that excess neutrons only produce attractive forces and this is sufficient enough to build stability.

Q13.17Written answer

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.

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Worked solution

A

No.of

atoms

B

Time

Q13.18Written answer

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.

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Worked solution

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

Half-life of C¹⁴ = 5760 years

Using radioactive law,

R = Roe^-λt

t = 2391.20 year

Q13.19Written answer

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.

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Worked solution

λ = h/p and kinetic energy = potential energy

E = hc/ λ

Kinetic energy of an electron

KE = PE = hc/ λ = 10⁹ eV

Q13.20Written answer

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

\(\begin{array}{l}Na_{11}^{23}\end{array} \)

?

(b) Which nuclide out of the two mirror isobars have greater binding energy and why?

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Worked solution

a) From the question, we know that a nuclide 1 is to be the mirror isobar of nuclide 2

if

Z1 = N2 and Z2 = N1

Therefore, mirror isobar is Z2 = 12 – N1 and N2 = 23 -12 = 11 = Z1

b) Mg has greater binding energy than Na.

Long Answers

Q13.21Written answer

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is

radioactive nucleus decays into a nucleus

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?

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Worked solution

Let λ1 and λ2 be the disintegration constants for S38 and Cl38

dN1/dt = – λN1

dN2/dt = rate of decay of Cl38 + rate of formation of Gl38

dN2/dt = -λ2N2 + λ1N1

e ^λ2tdN2 + λ2N2e ^λ2t dt = λ1Noe^{(λ2 – λ1)}dt

Integrating the above equation we get,

t = 1.65 h

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