If the nucleus $$_{13}^{27}Al$$ has a nuclear radius of about $$3.6\,fm,$$ then $$_{52}^{125}Te$$ would have its radius approximately as
A.
$$6.0\,fm$$
B.
$$9.6\,fm$$
C.
$$12.0\,fm$$
D.
$$4.8\,fm$$
Answer :
$$6.0\,fm$$
Solution :
If $$R$$ is the radius of the nucleus, the corresponding volume $$\frac{4}{3}\pi {R^3}$$ has been found to be proportional to $$A.$$
This relationship is expressed in inverse form as $$R = {R_0}{A^{\frac{1}{3}}}$$
The value of $${R_0}$$ is $$1.2 \times {10^{ - 15}}m,\,{\text{i}}{\text{.e}}{\text{.}}\,1.2\,fm$$
Therefore, $$\frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{R_0}{{\left( {{A_{Al}}} \right)}^{\frac{1}{3}}}}}{{{R_0}{{\left( {{A_{Te}}} \right)}^{\frac{1}{3}}}}}$$
$$\eqalign{
& \frac{{{R_{Al}}}}{{{R_{Te}}}} = \frac{{{{\left( {{A_{Al}}} \right)}^{\frac{1}{3}}}}}{{{{\left( {{A_{Te}}} \right)}^{\frac{1}{3}}}}} = \frac{{{{\left( {27} \right)}^{\frac{1}{3}}}}}{{{{\left( {125} \right)}^{\frac{1}{3}}}}} = \frac{3}{5} \cr
& {\text{or}}\,\,{R_{Te}} = \frac{5}{3} \times {R_{Al}} = \frac{5}{3} \times 3.6 = 6\,fm \cr} $$
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is