1.
A reactor converts $$100\% $$ of given mass into energy and that it operates at a power of $$9 \times {10^7}watt.$$ The mass of the fuel consumed in $$30\,\min$$ in the reactor will be :
the reactor will conume $$\frac{{9 \times {{10}^7}}}{{9 \times {{10}^{16}}}} \times 1800\,kg$$
$$ = 18 \times {10^{ - 7}}kg$$
2.
A radioactive sample with a half-life of 1 month has the label : ‘Activity = 2 microcurie on 1-8-1991’. What would be its activity two months earlier ?
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.
3.
Atomic weight of boron is 10.81 and it has two isotopes $$_5^{10}B$$ and $$_5^{11}B.$$ Then, the ratio of atoms of $$_5^{10}B$$ and $$_5^{11}B$$ in nature would be
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
4.
It is found that if a neutron suffers an elastic collinear
collision with deuterium at rest, fractional loss of its energy
is $${P_d};$$ while for its similar collision with carbon nucleus at rest, fractional loss of energy is $${P_c}.$$ The values of $${P_d}$$ and $${P_c}$$ are respectively:
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 $$ = \frac{{{\text{volume of atom}}}}{{{\text{volume of nucleus}}}}$$
$$\eqalign{
& = \frac{{\frac{4}{3}\pi r_1^3}}{{\frac{4}{3}\pi r_2^3}} \cr
& = {\left( {\frac{{{{10}^{ - 10}}}}{{{{10}^{ - 15}}}}} \right)^3} \cr
& = {10^{15}} \cr} $$
6.
If the nuclear force between two protons, two neutrons and between proton and neutron is denoted by $${F_{app}},{F_{nn}}$$ and $${F_{pn}}$$ respectively, then
A.
$${F_{pp}} \approx {F_{nn}} \approx {F_{pn}}$$
B.
$${F_{pp}} \ne {F_{nn}}\,{\text{and}}\,{F_{pp}} = {F_{nn}}$$
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.
8.
Energy released in the fission of a single $$_{92}{U^{235}}$$ nucleus is $$200\,MeV.$$ The fission rate of a $$_{92}{U^{235}}$$ filled reactor operating at a power level of $$5\,W$$ is
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.
10.
When an $$\alpha $$-particle of mass $$m$$ moving with velocity $$v$$ bombards on a heavy nucleus of charge $$Ze,$$ its distance of closest approach from the nucleus depends on $$m$$ as
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.