If an orbital electron of the hydrogen atom jumps from the groud state to a higher energy state, its orbital value where its velcoity is reduced to half its initial value.. If the radius of the electron orbit in the ground state is `r`, then the radius of the new orbit would be:

Options

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

Answer

B

Explanation

Since speed reduces to half, `KE` reduced to
`(1)/(4)th rArrn=2`
`mvr=(nh)/(2pi)`
`mv_(0)=r=1.(h)/(2pi)`........(I)
`m(v_(0))/(2)r=2.(h)/(2pi)`.....(II)
`r'=4.r`