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Q.

If the radius of a nucleus ${}^{256}X$ is is 8 fermi, the radius of ${}^{4}H e$ nucleus will be:

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a

1 fermi

b

2 fermi

c

3 fermi

d

4 fermi

answer is B.

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Detailed Solution

$Since R = R_{0} {(A)}^{\frac{1}{3}},$

Therefore,

$\frac{R_{2}}{R_{1}} = {(\frac{A_{2}}{A_{1}})}^{\frac{1}{3}} = {(\frac{4}{256})}^{\frac{1}{3}} = \frac{1}{4}$

$R_{2} = \frac{R_{1}}{4} = \frac{8}{4} = 2 fermi$

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