The de-Broglie wavelength of an electron moving in the nth Bohr orbit of radius ris given by

Options

A. `(nr)/(2pi)`
B. `(2pir)/(n)`
C. `(nr)/(pi)`
D. `npir`

Answer

B

Explanation

The de broglie wavelength of an electron is given by
`lamda = (h)/(mv) " or" mv = h/lamda " " …(1)`
and for the electron moving in the nth orbit,
The angular momentum `Iomega = (nh)/(2pi)`
`therefore mr^2 xx v/r = (nh)/(2pi) therefore mvr = (nh)/(2pi)`
`therefore mv = (nh)/(2pir) " "...(2)`
`therefore` From (1) and (2), `h/lamda = (nh)/(2pir)`
`therefore lamda =(2pir)/(n)`