The radius of first Bohr orbit is x, then de-Broglie wavelength of electron in 3rd orbit is nearly

Options

A. ` 3pi x`
B. ` 6 pi x`
C. ` 9x//2`
D. ` x//2`

Answer

B

Explanation

`lambda = h/( mu) = ( h.n.h)/( m 2pie^2) ( :. U = ( 2 pi e^2)/( nh))`
` = ( nh^2 . 4 pi^2 m. r_1)/( m. 2 pi . H^2) ( :. R_1 h_2/( 4 pi^2 me^2))`
`lambda = 3 xx 2 pi r_1 = 6 pi x ( n =3)`.