The time taken by a particle moving with velocity $c = 10^{6} m s^{- 1}$ in crossing a nucleus will approximately be
$10^{- 12}$ second
$10^{- 8}$ second
$10^{- 17}$ second
$10^{- 21}$ second

Question

The time taken by a particle moving with velocity c - 10 -6- ms-1- in crossing a nucleus will approximately be10-12- second10-8- second10-17- second10-21- second

Toppr Admin · Accepted Answer

Radius of a nucleus is given by the relation
$R = R_{0} (A^{\frac{1}{3}})$
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'.

The time taken by a particle moving with velocity c=106ms−1 in crossing a nucleus will approximately be10−12 second10−8 second10−17 second10−21 second

Radius of a nucleus is given by the relationR=R0(A13)where R0=1.2×10−15mHence, radius of the nucleus will be in order r=10−15m and so will be its diameter.Given, speed is c=106m/sTherefore, time taken to cross the nucleus is t=rc=10−15106=10−21 second approximately.Answer is 'D'.

The time taken by a particle moving with velocity $c = 10^{6} m s^{- 1}$ in crossing a nucleus will approximately be
$10^{- 12}$ second
$10^{- 8}$ second
$10^{- 17}$ second
$10^{- 21}$ second