The radius of electron orbit and the speed of electron in the ground state of hydrogen atom is `5.30xx10^(-11)m` and `22xx10^(6)m s^(-1)` respectively, then the orbital period of this electron in second excited state will be

Options

A. `1.21xx10^(-14)s`
B. `1.21xx10^(-12)s`
C. `1.21xx10^(-10)s`
D. `1.21xx10^(-15)s`

Answer

D

Explanation

Here, `r_1=5.30xx10^(-11)m,v_1=2.2xx10^6ms^(-1)`
In the second excited state, `r_n=n^2r_1,v_n=(v_1)/(n)`
`therefore r_2=4r_1=4xx5.30xx10^(-11)m=2.12xx10^(-10)m` and `v_2=(v_1)/(2)=(2.2xx10^(6))/(2)ms^(-1)=1.1xx10^(6)ms^(-1)`