Question

In hydrogen atom an electron revolves in a circular path. If the radius of its path is $0.53A^o$ and it makes $7\times {10}^{15}$ revolution per second, its angular momentum about proton is

• A. $11.2\times {10}^{-35}$ Joule/sec
• B. $11.2\times {10}^{-34}$ Joule/sec
• C. $11.2\times {10}^{-33}$ Joule/sec
• D. $11.2\times {10}^{-32}$ Joule/sec

Answer 1

The correct option is A $11.2\times {10}^{-35}$ Joule/sec

Given,
$r=0.53\times {10}^{-10}$
$\omega =7\times {10}^{15}$ rev per sec=7(2$\pi$ ) ${10}^{15}$ rad per sec
angular momentum about proton= $m{r}^2\omega$
=$9.1\ast {10}^{-31}(0.53\ast {10}^{-10}{)}^{2}7(2\pi ){10}^{15}$
=$11.2\times {10}^{-35}$.
Therefore the answer is option A