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 also four fold
B. K two fold, U also two fold
C. K four fold, U two fold
D. K two fold, U four fold

Answer

A

Explanation

K.E. of an electron in nth orbit : `K_(n) prop 1/(n^(2))`, and
P.E. of an electron in nth orbit : `U_(n) prop 1/(n^(2))`.
`:.` When an electron passes form state `n=2` to `n=1`,
`(K_(2))/(K_(1))=(1^(2))/(2^(2))=1/4` or `K_(1)=4K_(2)`
`(U_(2))/(U_(1))=(1^(2))/(2^(2))=1/4` or `U_(1)=4U_(2)`