An electron is revolving in an orbit of radius `2Å` with a speed of `4 xx 10^(5) m//s`. The angular momentum of the electron is `[M_(e) 9=xx10^(-31) kg]`.

Options

A. `2xx10^(-35) kg m^(2) s^(-1)`
B. `72xx10^(-36)kg m^(2) s^(-1)`
C. `7.2xx10^(-34)kg m^(2) s^(-1)`
D. `0.72xx10^(-37) kg m^(2) s^(-1)`

Answer

B

Explanation

Angular momentum `L=mv xx r=9xx10^(-10)xx2xx10^(-10)=72xx10^(-36) kg m^(2) s^(-1)`