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

`lambda =h/(mv)=(hr)/(mv r)=(hr)/((nh)/(2pi))`
`lambda =(2pi r)/n`
`:. nlambda =2 pi r`
`lambda =2 pi r" "(n=1)`