The de-Broglie wavelength of an electron in 4th orbit is (where, r=radius of 1st orbit)

Options

A. `2 pi r`
B. `4 pi r`
C. `(pi r)/2`
D. `16 pi r`

Answer

C

Explanation

`n=4, lambda =?`
According to de-Broglie,
`lambda=h/(mv)=(hr)/(mvr)`
`lambda=(hr)/((nh)/(2pi))" "( :' mvr=(nh)/(2pi))`
`lambda=(2pi r)/n`
i.e., `n lambda=2 pi r`
For `n=4`
`lambda =(2pi r)/n=(2pi r)/4`
`lambda=(pi r)/2`.