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

Options

A. `2pi r`
B. `4 pi r`
C. `8pi r`
D. `16 pi r`

Answer

C

Explanation

Radius of nth orbit can be given as `r_(n)=n^(2)r`
where, r is the radius of 1st orbit.
If `lambda` is the de-Broglie wavelength of an electron while revolving in nth orbit of radius `r_(n)`, then `2pi r_(n)=n lambda`
`or lambda=(2pi r_(n))/(n)`
For 4th orbit, n = 4
`therefore lambda =(2pi r_(4))/(4)=(2pi (4)^(2)r)/(4)=8 pi r`