Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

If the radius of the ${}_{13}^{27}\mathrm{Al}$ nucleus is taken to be $R_{\mathrm{Al}}$ , then the radius of ${}_{53}^{125}\mathrm{Te}$ nucleus is nearly:

$(53/13)^{1/3} R_{\mathrm{Al}}$

$\frac{5}{3} R_{\mathrm{Al}}$

$\frac{3}{5} R_{\mathrm{Al}}$

$(13/53)^{1/3} R_{\mathrm{Al}}$

Step-by-Step Solution

Formula: The radius $R$ of a nucleus is related to its mass number $A$ by the relationship $R = R_0 A^{1/3}$ , where $R_0$ is a constant .
Identify Mass Numbers:

For Aluminum ( ${}^{27}_{13}\mathrm{Al}$ ), the mass number $A_1 = 27$ .
For Tellurium ( ${}^{125}_{53}\mathrm{Te}$ ), the mass number $A_2 = 125$ .

Set up Ratio: $\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \frac{R_0 (125)^{1/3}}{R_0 (27)^{1/3}}$ $\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \left(\frac{125}{27}\right)^{1/3}$
Calculate:

Cube root of 125 is 5 ( $5^3 = 125$ ).
Cube root of 27 is 3 ( $3^3 = 27$ ). $\frac{R_{\mathrm{Te}}}{R_{\mathrm{Al}}} = \frac{5}{3}$ $R_{\mathrm{Te}} = \frac{5}{3} R_{\mathrm{Al}}$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started