For the Bohr's first orbit of circumference ` 2pi r` , the de - Broglie wavelength of revolving electron will be

Options

A. ` 2 pi r`
B. `pi r`
C. `(1)/( 2 pi r)`
D. `(1)/( 4 pi r)`

Answer

A

Explanation

`mvr = ( nh)/( 2 pi)`According to Bohr's theory
`rArr 2 pi r = n ((h)/( mv)) = n lambda ` for `n = 1 , lambda = 2 pi r`