If the radius of first orbit of H-atom is `a_(0)`, then de-Broglie wavelength of electron in `4^(th)` orbit is

Options

A. `8pi a_(0)`
B. `(a_(0))/(4)`
C. `16a_(0)`
D. `2pi a_(0)`

Answer

A

Explanation

`r_(n)=a_(0)xxn^(2)`
`r_(4)=a_(0)xx(4)^(2)=16a_(0)`
`mvr =(nh)/(2pi), mv=(4h)/(2pixx16 a_(0))`
`lamda=(h)/(mv)=(h)/(h//8pi a_(0))=8 pi a_(0)`