Question

The ratio of the surface area of the nuclei $T{e_{52}}^{125}$ to that of $A{l_{13}}^{27}$ is:
$(a){\text{ }}\dfrac{5}{3} \\\ (b){\text{ }}\dfrac{{125}}{{17}} \\\ (c){\text{ }}\dfrac{1}{4} \\\ (d){\text{ }}\dfrac{{25}}{9} \\\ (e){\text{ }}\dfrac{3}{5} \\\$

Answer 1

- Hint: In this question use the mass numbers of both the nuclei that is for $T{e_{52}}^{125}$ it is 125 and for ${}_{13}A{l^{27}}$it is 27. Use the direct relationship between the radius of the nuclei with that of the atomic mass that is $R = {R_o}{\left( A \right)^{\dfrac{1}{3}}}$ , then use the concept that surface area is directly proportional to the square of radii of the nucleus, this will help getting the right ratio of the surface area.

Complete step-by-step solution -

As we know the radius of the nucleus is given as
$R = {R_o}{\left( A \right)^{\dfrac{1}{3}}}$,
Where, R = radius of the nucleus
${R_o}$ = empirical constant = $1.1 \times {10^{ - 15}}$m
A = mass number of the nucleus.
So the radius of the nucleus is directly proportional to the cube root of the mass of the nuclei.
$\Rightarrow R \propto {\left( A \right)^{\dfrac{1}{3}}}$
So the ratio of the radius of the nuclei ${}_{52}T{e^{125}},{}_{13}A{l^{27}}$ is
As the mass number of the Te is 125 and the mass number of the Al is 27.
And 52 and 13 are the number of protons available in Te and Al.
Number of neutrons present in the Te and Al is the difference of the mass number and the protons of Te and Al respectively.
Where, Te is tellurium and Al is aluminum.
$\Rightarrow \dfrac{{{R_{Te}}}}{{{R_{Al}}}} = {\left( {\dfrac{{125}}{{27}}} \right)^{\dfrac{1}{3}}}$.................... (1)
Now the surface area of the nuclei is directly proportional to the square of the radius of the nuclei therefore,
$\Rightarrow \dfrac{{{A_{Te}}}}{{{A_{Al}}}} = {\left( {\dfrac{{{R_{Te}}}}{{{R_{Al}}}}} \right)^2}$
Now from equation (1) we have,
$\Rightarrow \dfrac{{{A_{Te}}}}{{{A_{Al}}}} = {\left( {\dfrac{{125}}{{27}}} \right)^{\dfrac{2}{3}}}$
Now simply this we have,
$\Rightarrow \dfrac{{{A_{Te}}}}{{{A_{Al}}}} = {\left( {{{\left( {\dfrac{5}{3}} \right)}^3}} \right)^{\dfrac{2}{3}}} = {\left( {\dfrac{5}{3}} \right)^2} = \dfrac{{25}}{9}$
So this is the required ratio of the surface area of the nuclei ${}_{52}T{e^{125}}$ to that of${}_{13}A{l^{27}}$.
So this is the required answer.

Hence option (D) is the correct answer.

Note – The trick point here was that the nuclei is basically in the form of a circular outline and thus the surface area is somewhat similar to the area of the circle which is considered as $\pi {r^2}$where r is the radius of the nuclei. This is the main reason why the surface area of the nuclei is taken as directly proportional to the square of the radius.