if speed of electron is ground state energy level is `2.2xx10^(6)ms^(-1)`, then its speed in fourth excited state will be

Options

A. `6.8xx10^(6)ms^(-1)`
B. `8.8xx10^(5)ms^(-1)`
C. `5.5xx10^(5)ms^(-1)`
D. `5.5xx10^(6)ms^(-1)`

Answer

C

Explanation

According to bohr's model `v=(2Ke^(2)Z)/(nh)or vprop(1)/(n)therefore(v_A)/(v_B)=(n_B)/(n_A)`
Here,`v_A=2.2xx10^(6)ms^(-1)`,`n_A=1,n_B=4`
`therefore v_B=v_Axx(n_A)/(n_B)=2.2xx10^(6)xx(1)/(4)=0.55xx10^(6)`
`=5.5xx10^(5)ms^(-1)`