Question

If radius of the ${}_{13}^{27}Al$ nucleus is estimated to be $3.6$ fermi, then the radius of ${}_{52}^{125}Te$ nucleus be nearly.

• A. $6fermi$
• B. $8fermi$
• C. $4fermi$
• D. $5fermi$

Answer 1

The correct option is A $6fermi$

Radius of a nucleus, R = R${}_0$A${}^{1/3}$ where R${}_0$ is a contant and A is mass no.

For Al, R = R${}_0$(27)${}^{1/3}$

For Te, R' = R${}_0$(125)${}^{1/3}$

On solving both the above equations, R' = 6 fermi