According to Bohr's theory, which of the following transition is H-atom will gie the least energetic photon-

Options

A. `n=6` to n=1
B. n=5 to n=4
C. n=6 to n=5
D. n=5 to n=3

Answer

C

Explanation

`DeltaE=E_(n_(2))-E_(n_(1))=(2pi^(2)me^(4))/(h^(2))[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`therefore Delta Eprop[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
Hence, transition from n=6 to n=5 will give rise to the least energetic photon.