Atoms or Nuclear Fission and Fusion MCQ Questions & Answers in Modern Physics | Physics

The binding energy curve has a broad maximum in the range $$A = 30$$  to $$A = 120$$  corresponding to average binding energy per nucleon $$= 8\,MeV.$$   The peak value of the maximum is $$8.8\,MeV/N$$   for $$_{26}F{e^{56}}.$$

Stellar energy is the energy obtained continuously from the sun and the stars. It is estimated that the sun has been radiating $$3.8 \times {10^{26}}J$$   of energy per second for billions of years. Bethe postulated that the interior of the sun and stars provide conditions for the fusion of hydrogen nuclei to form helium nuclei with the release of heavy amount of energy. Hence, solar energy is due to fusion reaction.

The penetrating power of radiation is directly proportional to the energy of its photon.
Energy of photon $$ = \frac{{hc}}{\lambda } \propto \frac{1}{\lambda }\left( {{\text{wavelength}}} \right)$$
∴ Penetrating power $$ \propto \frac{1}{\lambda }\left( {{\text{wavelength}}} \right)$$
Since, $$\lambda $$ is minimum for $$\gamma $$-rays, so penetrating power is maximum for $$\gamma $$-rays.

The nuclei which have same number of neutrons but different atomic number and mass number are known as isotones. In choice (A) nuclei of $$_{34}S{e^{74}}$$  and $$_{31}G{a^{71}}$$  are isotones as
$$A - Z = 74 - 34 = 71 - 31 = 40$$

$$_0^1n \to _1^1H{ + _{ - 1}}{e^0} + \bar v + Q$$
The mass defect during the process
$$\eqalign{ & \Delta m = {m_n} - {m_H} - {m_e} = 1.6725 \times {10^{ - 27}} - \left( {1.6725 \times {{10}^{ - 27}} + 9 \times {{10}^{ - 31}}kg} \right) \cr & = - 9 \times {10^{ - 31}}kg \cr} $$
The energy released during the process
$$\eqalign{ & E = \Delta m{c^2} \cr & E = 9 \times {10^{ - 31}} \times 9 \times {10^{16}} = 81 \times {10^{ - 15}}Joules \cr & E = \frac{{81 \times {{10}^{ - 15}}}}{{1.6 \times {{10}^{ - 19}}}} \cr & = 0.511\,MeV \cr} $$

Number of atoms in $$2kg$$  fuel
$$ = \frac{2}{{235}} \times 6.023 \times {10^{26}} = 5.12 \times {10^{24}}$$
number of fission per second
$$ = \frac{{5.12 \times {{10}^{24}}}}{{30 \times 24 \times 60 \times 60}} = 1.978 \times {10^{18}}$$
Energy released per fission
$$ = 200\,MeV = 200 \times 1.6 \times {10^{ - 13}} = 3.2 \times {10^{ - 11}}J$$
Power output $$ = 3.2 \times {10^{ - 11}} \times 1.978 \times {10^{18}}$$
$$ = 63.28 \times {10^6}W = 63.28\,MW$$

The binding energy per nucleon for the middle nuclides (from $$A = 20$$  to $$A = 56$$  ) is maximum. Hence, these are more stable.
As the mass number increases, the binding energy per nucleon gradually decreases and ultimately binding energy per nucleon of heavy nuclides (such as uranium etc.) is comparatively low. Hence, these nuclides are relatively unstable. So, they can be fissioned easily.

Mass number = No. of protons + No. of neutrons
For example, in case of hydrogen
Number of neutrons = 0
Thus, mass number = atomic number
Hence, sometimes the atomic number is equal to the mass number.

$$\eqalign{ & _1{H^2}{ + _1}{H^2}{ \to _2}H{e^4} + Q \cr & \Rightarrow \Delta m = m\left( {_2H{e^4}} \right) - 2m\left( {_1{H^2}} \right) \cr & \Rightarrow \Delta m = 4.0024 - 2\left( {2.0141} \right) \cr & \Rightarrow \Delta m = - 0.0258\,amu \cr & {\text{Since,}}\,Q = {c^2}\Delta m \cr & \Rightarrow Q = \left( {0.0258} \right)\left( {931.5} \right)MeV \cr & \Rightarrow Q = 24\,\,MeV. \cr} $$

In a nuclear fusion, when two light nuclei of different masses are combined to form a stable nucleus, then some mass is lost and appears in the form of energy, called the mass defect. So, the mass of resultant nucleus is always less than the sum of masses of initial nuclei i.e.,
$${m_3} < \left( {{m_1} + {m_2}} \right)$$