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. 4 Fermi
• B. 5 Fermi
• C. 6 Fermi
• D. 8 Fermi

Answer 1

Answer: (C)

6 Fermi

Answer 2

r µ A^1/3 Ž = $\left( \frac { \mathrm { A } _ { \mathrm { Te } } } { \mathrm { A } _ { \mathrm { Al } } } \right) ^ { 1 / 3 }$

= $\left( \frac { 125 } { 27 } \right) ^ { 1 / 3 }$

r_Te = $\frac { 5 } { 3 }$ × r_Al = $\frac { 5 } { 3 }$ × 3.6 = 6 fermi