According to de - Broglie , the de - Broglie wavelength for electron in an orbit of hydrogen atom is `10^(-9) m`. The principle quantum number for this electron is

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

C

Explanation

`2 pi r n = lambda rArr n = (l)/( 2 pi r) = ( 10^(-9))/(2 xx 3.14 xx 5.13 xx 10^(-11)) = 3`