In hydrogen atom , electron excites from ground state of higher energy state and its orbital velocity is reduced to `((1)/(3))^(rd)` of its initial value . The radius of the orbit in the ground state is R . The radius of the orbit in that higher energy state is ...

Options

A. 2 R
B. 3 R
C. 27 R
D. 9 R

Answer

D

Explanation

As per question ,
`(v_(h))/(v_(g)) = (1)/(3) " " … (i)`
where subscripts h and q denotes higher energy state and ground state .
Orbital velocity of electron in the `n^(th)` orbit is `v_(n) = (e^(2))/( 2 epsi_(0) nh) or v_(n) prop (1)/(n)`
For ground state , n = 1
`therefore (v_(h))/(v_(g)) = (1)/(n) " " ... (ii)`
Equating (i) and (ii) , we get n = 3
Radius of `n^th` orbit is
`r_(n) = (n^(2) h^(2) epsi_(0))/(pi e^(2) m)`
or `r_(n) prop n^(2)`
`therefore (r_(3))/(r_(1)) = ((3)^(2))/((1)^(2)) = 9`
`r_(3) = 9 r_(1) = 9 R " " (because r_(1) = R` (Given))