Question

The time taken by a particle moving with velocity $c={10}^{6}m{s}^{-1}$ in crossing a nucleus will approximately be

• A. ${10}^{-12}$ second
• B. ${10}^{-8}$ second
• C. ${10}^{-17}$ second
• D. ${10}^{-21}$ second

Answer 1

The correct option is D ${10}^{-21}$ second

Radius of a nucleus is given by the relation
$R=R_0\left(A^{\frac{1}{3}}\right)$
where $R_0=1.2\times {10}^{-15}m$
Hence, radius of the nucleus will be in order $r={10}^{-15}m$ and so will be its diameter.
Given, speed is $c={10}^{6}m/s$
Therefore, time taken to cross the nucleus is $t=\frac{r}{c}=\frac{{10}^{-15}}{{10}^6}={10}^{-21}$ second approximately.
Answer is 'D'.