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

the reactor will conume $$\frac{{9 \times {{10}^7}}}{{9 \times {{10}^{16}}}} \times 1800\,kg$$
$$ = 18 \times {10^{ - 7}}kg$$

The activity of a radioactive substance is defined as the rate at which the nuclei of its atoms in the sample disintegrate. In two half-lives, the activity becomes one-fourth. Two months is 2 half-life period. The activity, two months earlier was
$$2 \times {2^2} = {\text{8}}\,{\text{microcurie}}{\text{.}}$$
NOTE
The activity of a radioactive sample is called one curie, if it undergoes $$3.7 \times {10^{10}}$$   disintegrations per second.

Let $${n_1}$$ and $${n_2}$$ be the number of atoms in $$_5^{10}B$$  and $$_5^{11}B$$  isotopes.
Atomic weight $$ = \frac{{{n_1} \times \left( {{\text{At}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of }}_5^{10}\;B} \right) + {n_2} \times \left( {{\text{At}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of }}_5^{11}\;B} \right)}}{{{n_1} + {n_2}}}$$
$$\eqalign{ & {\text{or}}\,\,10.81 = \frac{{{n_1} \times 10 + {n_2} \times 11}}{{{n_1} + {n_2}}} \cr & {\text{or}}\,\,10.81\,{n_1} + 10.81\,{n_2} = 10\,{n_1} + 11\,{n_2} \cr & {\text{or}}\,\,0.81\,{n_1} = 0.19\,{n_2} \cr & {\text{or}}\,\,\frac{{{n_1}}}{{{n_2}}} = \frac{{0.19}}{{0.81}} = \frac{{19}}{{81}} \cr} $$
NOTE
Atomic weight of an atom having two or more isotopes is the average of the total weight of two of more isotopes

For collision of neutron with deuterium:

Applying conservation of momentum :
$$\eqalign{ & mv + 0 = m{v_1} + 2m{v_2}\,......\left( {\text{i}} \right) \cr & {v_2} - {v_1} = v\,......\left( {{\text{ii}}} \right) \cr} $$
$$\because $$ Collision is elastic, $$e = 1$$
From eqn (i) and eqn (ii) $${v_1} = - \frac{v}{3}$$
$${P_d} = \frac{{\frac{1}{2}m{v^2} - \frac{1}{2}mv_1^2}}{{\frac{1}{2}m{v^2}}} = \frac{8}{9} = 0.89$$
Now, For collision of neutron with carbon nucleus

Applying Conservation of momentum
$$\eqalign{ & mv + 0 = m{v_1} + 12m{v_2}\,......\left( {{\text{iii}}} \right) \cr & v = {v_2} - {v_1}\,......\left( {{\text{iv}}} \right) \cr} $$
From eqn (iii) and eqn (iv)
$$\eqalign{ & {v_1} = - \frac{{11}}{{13}}v \cr & {P_c} = \frac{{\frac{1}{2}m{v^2} - \frac{1}{2}m{{\left( {\frac{{11}}{{13}}v} \right)}^2}}}{{\frac{1}{2}m{v^2}}} = \frac{{48}}{{169}} \approx 0.28 \cr} $$

Order of Radius of atom $$ \approx {10^{ - 10}}m$$
Order of Radius of nucleus $$ \approx {10^{ - 15}}m$$
Ratio of volume of atom to volume of nucleus $$ = \frac{{{\text{volume of atom}}}}{{{\text{volume of nucleus}}}}$$
$$\eqalign{ & = \frac{{\frac{4}{3}\pi r_1^3}}{{\frac{4}{3}\pi r_2^3}} \cr & = {\left( {\frac{{{{10}^{ - 10}}}}{{{{10}^{ - 15}}}}} \right)^3} \cr & = {10^{15}} \cr} $$

Nuclear forces act between a pair of neutrons, a pair of protons and also between a neutron-proton pair, with the same strength. This shows that nuclear forces are independent of charge.

Out of the given choices, X-rays and $$\gamma $$-rays are electromagnetic waves, so they have no charge. $$\beta $$-particles are negatively charged particles and are fast moving electrons. Alpha $$\left( \alpha \right)$$  particles have positive charge and is a nucleus of helium.

Fission rate = $$\frac{{{\text{total nuclear power}}}}{{{\text{energy produced/fission}}}}$$
Here, total nuclear power $$= 5\,W$$
Energy released per fission $$= 200\,MeV$$
∴ Fission rate $$ = \frac{5}{{200\,MeV}}$$
$$\eqalign{ & = \frac{5}{{200 \times 1.6 \times {{10}^{ - 13}}}}\,\,\left[ {\because 1\,MeV = 1.6 \times {{10}^{ - 13}}J} \right] \cr & = 1.56 \times {10^{11}}{s^{ - 1}} \cr} $$

Fission of $$_{92}^{235}U$$  is possible by only slow neutrons. To produce slow neutrons, one has to use moderators so that fission is possible.

When an $$\alpha $$-particle of mass $$m$$ moving with velocity $$v$$ bombards on a heavy nucleus of charge $$Ze,$$  then there will be no loss of energy as in this case, initial kinetic energy of $$\alpha $$-particle = potential energy of $$\alpha $$-particle at closest approach.
$$\eqalign{ & \Rightarrow \frac{1}{2}m{v^2} = \frac{{2Z{e^2}}}{{4\pi {\varepsilon _0}{r_0}}} \cr & \Rightarrow \boxed{{r_0} \propto \frac{1}{m}} \cr} $$
This is the required distance of closest approach to $$\alpha $$-particle from the nucleus.