If an electron in hydrogen atom jumps from third orbit to second orbit, the frequency of the emitted radiation is given by (c is speed of light )

Options

A. `4.5 xx10^(14)` Hz
B. `5.4xx10^(14)` Hz
C. `4.5xx10^(-14)` Hz
D. `5.5xx10^(14)` Hz

Answer

A

Explanation

`hv=E_(n)-E_(p)`
`=E_(3)-E_(2)`
`=E_(1)/9-E_(1)/4`
`=-13.6/9+13.6/4`
`=-1.51+3.4`
`=1.89 eV`
`:. v=(1.89xx1.6xx10^(-19))/(6.63xx10^(-34))=4.5xx10^(14) Hz`