Question

Radius of nucleus varies as $R=R_0(A{)}^{1/3}$, where $R_0=1.3$ Fermi. What is the volume of $B{e}^8$ nucleus (approx) [$A$ = atomic mass] ?

• A. $7\times {10}^{-38}cc$
• B. $7\times {10}^{-29}cc$
• C. $7\times {10}^{-45}cc$
• D. None of these

Answer 1

The correct option is A $7\times {10}^{-38}cc$

Radius of nucleus is $R=1.3\times (8{)}^{1/3}Fermi=2.6Fermi=2.6\times {10}^{-15}meter=2.6\times {10}^{-13}cm$

The nucleus is supposed to be sphere so $volume=\left(4\pi /3\right)R^3=4.1866\times (2.6\times {10}^{-13}{)}^3\text{cubic centimeter(cc)}$

or Volume $=7.36\times {10}^{-38}cc$

Option A is correct.