An electron of mass `9xx10^(-31) kg` revolues in a circle of radius `0.53 A` around the nucles of hydrogen with a velocity of `2.2xx10^(6) ms^(-1)`. Show that angular momentum of electron is `h//2 pi`, where `h` is plack's constant.

Explanation

Here, `m = 9 xx10^(-31) kg, r = 0.53 A = 0.53 xx 10^(-10)m, upsilon = 2.2 xx 10^(6) m//s`
Angular momentum, `L = m upsilon r = 9 xx 10^(-31) xx 2.2 xx 10^(6) xx 0.53 xx 10^(-10) = 1.0494 xx 10^(-34) Js`.
Also, `(h)/(2pi) = (6.6 xx 10^(-34))/(2xx22//7) = 1.05 xx 10^(-34) Js`. Hence, `L~~ h//2 pi`