Question 1

Fusion reaction takes place at high temperature because

Accepted Answer

kinetic energy is high enough to overcome the coulomb repulsion between nuclei

Question 2

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

Accepted Answer

elastic collision of neutrons with heavy nuclei will not slow them down.

Question 3

When a nucleus in an atom undergoes a radioactive decay, the electronic energy levels of the atom

Accepted Answer

change for α and β radioactivity but not for gamma-radioactivity.

Question 4

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,

Accepted Answer

the containers will in general have different numbers of the atoms of the material but their average will be close to 5000.

Question 5

Potential energy of a pair of of nucleons as a function of their separation is minimum at a distance of about

Accepted Answer

0.8 fm

Question 6

Which of the following statements is true for nuclear forces?

Accepted Answer

They are short-range forces

Question 7

Heavy stable nuclei have more neutrons than protons. This is because of the fact that

Accepted Answer

electrostatic forces between protons are repulsive

Question 8

Nuclides with same neutron number N but different atomic Number Z, are called

Accepted Answer

Isotones

Question 9

1 Obtain the binding energy (in MeV) of a nitrogen nucleus $^{14}_{7} \mathrm{N}$, given $m\left(^{14}_{7} \mathrm{N}\right) = 14.00307 \mathrm{u}$

Accepted Answer

Given: Mass of nitrogen nucleus $^{14}_{7} \mathrm{N} = 14.00307\,u$

Number of protons, $Z = 7$

Number of neutrons, $N = 14 - 7 = 7$

Mass of proton $m_p = 1.007825\,u$
Mass of neutron $m_n = 1.008665\,u$

Calculate the mass defect $\Delta m$:
$$
\Delta m = Z m_p + N m_n - m(^{14}_{7}N) = 7 \times 1.007825 + 7 \times 1.008665 - 14.00307
$$
$$
= 7.054775 + 7.060655 - 14.00307 = 14.11543 - 14.00307 = 0.11236\,u
$$

Convert mass defect to energy (binding energy):
$$
E = \Delta m \times 931.5\, \mathrm{MeV} = 0.11236 \times 931.5 = 104.66\, \mathrm{MeV}
$$

Therefore, the binding energy of $^{14}_{7} \mathrm{N}$ nucleus is approximately 104.7 MeV. — Step 1: Calculate total mass of free nucleons (protons + neutrons).
Step 2: Calculate mass defect $\Delta m$ by subtracting actual nuclear mass.
Step 3: Convert mass defect in atomic mass units to energy using $1\,u = 931.5\,\mathrm{MeV}$.
Step 4: Result is the binding energy of the nucleus.

Question 10

2 Obtain the binding energy of the nuclei $^{56}_{26} \mathrm{Fe}$ and $^{209}_{83} \mathrm{Bi}$ in units of MeV from the following data:

$m\left(^{56}_{26} \mathrm{Fe}\right) = 55.934939 \mathrm{u} \quad m\left(^{209}_{83} \mathrm{Bi}\right) = 208.980388 \mathrm{u}$

Accepted Answer

Given:
$m(^{56}_{26}Fe) = 55.934939\,u$, $Z=26$, $N=56-26=30$
$m(^{209}_{83}Bi) = 208.980388\,u$, $Z=83$, $N=209-83=126$

Mass of proton $m_p = 1.007825\,u$
Mass of neutron $m_n = 1.008665\,u$

For $^{56}_{26}Fe$:
$$
\Delta m = Z m_p + N m_n - m(^{56}_{26}Fe) = 26 \times 1.007825 + 30 \times 1.008665 - 55.934939
$$
$$
= 26.20345 + 30.25995 - 55.934939 = 56.4634 - 55.934939 = 0.528461\,u
$$
Binding energy:
$$
E = 0.528461 \times 931.5 = 492.1\, \mathrm{MeV}
$$

For $^{209}_{83}Bi$:
$$
\Delta m = 83 \times 1.007825 + 126 \times 1.008665 - 208.980388
$$
$$
= 83.649475 + 127.09179 - 208.980388 = 210.741265 - 208.980388 = 1.760877\,u
$$
Binding energy:
$$
E = 1.760877 \times 931.5 = 1640.6\, \mathrm{MeV}
$$

Therefore, binding energies are approximately 492.1 MeV for $^{56}_{26}Fe$ and 1640.6 MeV for $^{209}_{83}Bi$. — Calculate mass defect for each nucleus by subtracting actual nuclear mass from sum of free nucleons' masses. Convert mass defect to energy using $1u = 931.5 MeV$.

Question 11

3 A given coin has a mass of $3.0 \mathrm{g}$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of $^{63}_{29} \mathrm{Cu}$ atoms (of mass 62.92960 u).

Accepted Answer

Given:
Mass of coin $m = 3.0\,g = 3.0 \times 10^{-3} kg$
Atomic mass of copper $^{63}_{29}Cu = 62.92960\,u$

Number of moles of copper atoms in coin:
$$
\text{moles} = \frac{3.0\times10^{-3}}{62.92960 \times 1.6605 \times 10^{-27} \times 10^{3}} 
$$
But better to use molar mass in grams:
Molar mass $M = 62.92960\,g/mol$
$$
\text{moles} = \frac{3.0}{62.92960} = 0.0477\,mol
$$

Number of atoms:
$$
N = 0.0477 \times 6.023 \times 10^{23} = 2.87 \times 10^{22} \text{ atoms}
$$

Number of nucleons per atom $A = 63$

Binding energy per nucleon for copper is approximately 8.7 MeV (typical value for mid-mass nuclei).

Total binding energy:
$$
E = N \times A \times 8.7\, \mathrm{MeV} = 2.87 \times 10^{22} \times 63 \times 8.7 = 1.57 \times 10^{25} \mathrm{MeV}
$$

Convert MeV to Joules:
$$
1 \mathrm{MeV} = 1.6 \times 10^{-13} \mathrm{J} 
$$
$$
E = 1.57 \times 10^{25} \times 1.6 \times 10^{-13} = 2.51 \times 10^{12} \mathrm{J}
$$

Therefore, the nuclear energy required to separate all nucleons in the 3 g coin is approximately $2.5 \times 10^{12}$ Joules. — Step 1: Calculate moles of copper atoms in the coin.
Step 2: Calculate total number of atoms.
Step 3: Multiply by nucleons per atom and binding energy per nucleon to get total binding energy in MeV.
Step 4: Convert MeV to Joules.

Question 12

4 Obtain approximately the ratio of the nuclear radii of the gold isotope $^{107}_{79} \mathrm{Au}$ and the silver isotope $^{107}_{47} \mathrm{Ag}$.

Accepted Answer

Given:
Gold isotope $^{107}_{79}Au$, mass number $A_1 = 107$
Silver isotope $^{107}_{47}Ag$, mass number $A_2 = 107$

Nuclear radius formula:
$$
R = R_0 A^{1/3}
$$
Since both have same mass number $A=107$, their radii are:
$$
R_{Au} = R_0 (107)^{1/3}, \quad R_{Ag} = R_0 (107)^{1/3}
$$

Therefore,
$$
\frac{R_{Au}}{R_{Ag}} = 1
$$

Hence, the ratio of nuclear radii is approximately 1.

The nuclear radius depends only on mass number $A$, not on proton number $Z$. — Since nuclear radius depends on $A^{1/3}$ and both isotopes have same $A=107$, their radii are equal, so ratio is 1.

Question 13

5 The $Q$ value of a nuclear reaction $A + b \rightarrow C + d$ is defined by

$$
Q = \left[ m_{A} + m_{b} - m_{C} - m_{d} \right] c^{2}
$$

where the masses refer to the respective nuclei. Determine from the given data the $Q$-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) $^{1}_{1} \mathrm{H} + ^{3}_{1} \mathrm{H} \rightarrow ^{2}_{1} \mathrm{H} + ^{2}_{1} \mathrm{H}$

(ii) $^{12}_{6} \mathrm{C} + ^{12}_{6} \mathrm{C} \rightarrow ^{20}_{10} \mathrm{Ne} + ^{4}_{2} \mathrm{He}$

Atomic masses are given to be

$$
\begin{array}{l}
m\left(^{2}_{1} \mathrm{H}\right) = 2.014102 \mathrm{u} \
m\left(^{3}_{1} \mathrm{H}\right) = 3.016049 \mathrm{u} \
m\left(^{12}_{6} \mathrm{C}\right) = 12.000000 \mathrm{u} \
m\left(^{20}_{10} \mathrm{Ne}\right) = 19.992439 \mathrm{u} \
\end{array}
$$

Accepted Answer

(i) Reaction: $^{1}_{1}H + ^{3}_{1}H \rightarrow ^{2}_{1}H + ^{2}_{1}H$

Masses:
$$
m(^{1}_{1}H) = 1.007825\,u \quad (given in data section)
$$
$$
m(^{3}_{1}H) = 3.016049\,u
$$
$$
m(^{2}_{1}H) = 2.014102\,u
$$

Calculate $Q$:
$$
Q = [m_A + m_b - m_C - m_d] c^2
$$
$$
= [1.007825 + 3.016049 - 2.014102 - 2.014102] c^2
$$
$$
= (4.023874 - 4.028204) c^2 = -0.00433\,u \times 931.5\,MeV/u = -4.03\,MeV
$$

Since $Q < 0$, the reaction is endothermic.

(ii) Reaction: $^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{10}Ne + ^{4}_{2}He$

Masses:
$$
m(^{12}_{6}C) = 12.000000\,u
$$
$$
m(^{20}_{10}Ne) = 19.992439\,u
$$
$$
m(^{4}_{2}He) = 4.002603\,u \quad (given in data section)
$$

Calculate $Q$:
$$
Q = [12.000000 + 12.000000 - 19.992439 - 4.002603] c^2
$$
$$
= (24.000000 - 23.995042) c^2 = 0.004958\,u \times 931.5\,MeV/u = 4.62\,MeV
$$

Since $Q > 0$, the reaction is exothermic. — Calculate $Q$ by subtracting product masses from reactant masses and converting mass difference to energy using $1u = 931.5 MeV$. Negative $Q$ means energy absorbed (endothermic), positive $Q$ means energy released (exothermic).

Question 14

6 Suppose, we think of fission of a $^{56}_{26} \mathrm{Fe}$ nucleus into two equal fragments, $^{28}_{13} \mathrm{Al}$. Is the fission energetically possible? Argue by working out $Q$ of the process. Given $m\left(^{56}_{26} \mathrm{Fe}\right) = 55.93494 \mathrm{u}$ and $m\left(^{28}_{13} \mathrm{Al}\right) = 27.98191 \mathrm{u}$.

Accepted Answer

Given:
$m(^{56}_{26}Fe) = 55.93494\,u$
$m(^{28}_{13}Al) = 27.98191\,u$

Reaction:
$$
^{56}_{26}Fe \rightarrow 2 \times ^{28}_{13}Al
$$

Calculate $Q$:
$$
Q = [m(Fe) - 2 \times m(Al)] c^2 = [55.93494 - 2 \times 27.98191] c^2
$$
$$
= (55.93494 - 55.96382) c^2 = -0.02888\,u \times 931.5\,MeV/u = -26.9\,MeV
$$

Since $Q < 0$, the reaction is endothermic and not energetically possible spontaneously.

Therefore, fission of $^{56}_{26}Fe$ into two $^{28}_{13}Al$ nuclei is not energetically favorable. — Calculate mass difference between initial nucleus and sum of fission fragments. Negative $Q$ means energy input is required, so fission is not spontaneous.

Question 15

7 The fission properties of $^{239}_{94}\mathrm{Pu}$ are very similar to those of $^{235}_{92}\mathrm{U}$. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure $^{239}_{94}\mathrm{Pu}$ undergo fission?

Accepted Answer

Given:
Mass of plutonium sample $m = 1\,kg = 1000\,g$
Molar mass of $^{239}_{94}Pu = 239\,g/mol$
Energy released per fission = 180 MeV

Number of moles:
$$
\text{moles} = \frac{1000}{239} = 4.1841\,mol
$$

Number of atoms:
$$
N = 4.1841 \times 6.023 \times 10^{23} = 2.52 \times 10^{24} \text{ atoms}
$$

Total energy released:
$$
E = N \times 180 = 2.52 \times 10^{24} \times 180 = 4.54 \times 10^{26} \mathrm{MeV}
$$

Convert MeV to Joules:
$$
E = 4.54 \times 10^{26} \times 1.6 \times 10^{-13} = 7.26 \times 10^{13} \mathrm{J}
$$

Therefore, total energy released is approximately $7.3 \times 10^{13}$ Joules. — Step 1: Calculate number of moles in 1 kg of Pu.
Step 2: Calculate total number of atoms.
Step 3: Multiply by energy per fission to get total energy in MeV.
Step 4: Convert MeV to Joules.

Nuclei

Nuclei — Study Notes

13.1 INTRODUCTION

13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS

Discovery of Neutron

Practice Questions — Nuclei

All 6 Chapters in Physics Part-II