As the electron in the Bohr orbit is hydrogen atom passes from state `n = 2` to `n = 1` , the `KE (K) and PE (U)` change as

Options

A. K four-fold , U two - fold
B. K four-fold , U also four - fold
C. k two-fold, U also two - fold
D. K two - fold , U four - fold

Answer

B

Explanation

`For n =1, E_1 = -13.6 eV , K_1 = -E_1 = + 13.6 eV`
and `U_1 = 2E_1 = -13.6xx2 eV`
For n =2 , `E_2 = - (13.6)/(4) , K_2 = +(13.6)/(4)`
and `U_2 = -2xx(13.6)/(4) = - (13.6)/(2)`
Thus `E_1 /E_2 = 4, K_1/K_2 = 4`
and `U_1/U_2 = (-13.6xx2)/((-13.6)/(2)) = 4`
Thus both K.E. and P.E. become four times or four fold.