Question

Two nuclei have mass numbers in the ratio $ 27:125 $ , what is the ratio of their nuclear radii?

Answer 1

Hint: This question is related to nuclear chemistry and physics. The formula that will be applied here is the formulaAlkanesAlkanesAlkanes of the average radius of the nucleus with a nucleon. In the question mass number values are given in ratio just have to substitute the values $ 27:125 $ and by solving within 2-3 steps you will get the required answer.

Complete answer:
Given above is the mass number of two nuclei in ratio $ 27:125 $ we need to find the ratio of their nuclear radii. The formula that will be used here is the formula of average radius of the nucleus given as:
 $ R{\text{ }} = {\text{ }}{R_0}{A^{1/3}}\;\;\; $
And since we are given here with two nuclei we will write the formula for two nuclei as
 $ {R_1}{\text{ }} = {\text{ }}{R_0}{A_1}^{1/3}\;\;\; $ And $ {R_2}{\text{ }} = {\text{ }}{R_0}{A_2}^{1/3}\;\;\; $
When we take the ratio of the two it can be written as
 $ \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_0}}}{{{R_0}}} \times {\left( {\dfrac{{{A_1}}}{{{A_2}}}} \right)^{\dfrac{1}{3}}} $
From the above equation $ {A_1}{\text{ }}and{\text{ }}{A_2} $ are given in ratio as $ 27:125 $ and we need to find $ R_1{\text{ }}and{\text{ }}R_2. $ R0 being common in both can be cancelled out.
So after substituting the value the equation become:
 $ \dfrac{{{R_1}}}{{{R_2}}} = {\left[ {{{\left( {\dfrac{3}{5}} \right)}^3}} \right]^{\dfrac{1}{3}}} $
(In this step we have just taken the cube root of $ 27{\text{ }}and{\text{ }}125 $ )
Next just cancel out $ 3 $ from and we get the final answer.
 $ \dfrac{{{R_1}}}{{{R_2}}} = {\left[ {{{\left( {\dfrac{3}{5}} \right)}^{{3}}}} \right]^{\dfrac{1}{{{3}}}}} $
 $ \dfrac{{{R_1}}}{{{R_2}}} = {\left[ {\left( {\dfrac{3}{5}} \right)} \right]^1} $
This can be written as
 $ \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{3}{5} $
Thus we get $ R_1{\text{ }}and{\text{ }}R_2. $ as $ 3{\text{ }}and{\text{ }}5 $ . Therefore the ratio of their nuclear radii $ R_1:R_2 $ is $ 3:5 $ .

Note:
The equation used here is $ R{\text{ }} = {\text{ }}{R_0}{A^{1/3}}\;\;\; $ , (where R is the radius, A is the mass number of the nuclei and R0 is the constant whose value is given as $ 1.2{\text{ }} \times {\text{ }}{10^{ - 15}} $ meter or $ 1.2 $ fm where fm is fermi). The mass number is the total number of protons and neutrons in the nucleus.