The frequency of light emitted, when the electron makes transition from the level of principle quantum number n = 2 to the level with n = 1 is (Take, the ionization energy of hydrogen to be 13.6eV andh`=4xx 10 ^(-15)` eV -s )

Options

A. `2.55xx10^(15)Hz`
B. `1.7xx10^(15)Hz`
C. `3.4xx10^(15)Hz`
D. `5.1xx10^(15)Hz`

Answer

A

Explanation

As ionisation energy is 13.6 eV, so its negative will be the energy of atom in ground state (n=1)
hence `E_(1) =` - 13.6 eV
Energy in second orbit (n =1)
`E_(2)=-(13.6)/(2^(2))=- 3.4` eV
When electron jumps from n = 2 to n = 1, then energyn of photon emitted by atom
`DeltaE =E_(2)-E_(1)`
`= - 3.4 - (-13.6)`
`DeltaE = 10.2` eV
So, frequency (v),
hv = 10.2 eV
`v = (10.2 eV)/(h) = (10.2 eV)/(4xx10^(-15)eV-s)`
`v= 255xx10^(+15)Hz`