Calculate the wavelength of the radiations in nanometers emitted when an electron in hydrogen atom jumps from third orbit to the ground state. `(R_(H)=109677 cm^(-1))`

Options

A. 103 nm
B. 216 nm
C. 110 nm
D. 246 nm

Answer

A

Explanation

`R_(H)=109677 cm^(-1)`,
`n_(1)=1, n_(2)=3`
`bar(v)=109677[(1)/(1^(2))-(1)/(3^(2))]cm^(-1)`
`109677xx(8)/(9)cm^(-1)=97490.7 cm^(-1)`
wavelength `(lambda)=(1)/(bar(v))=(1)/(97490.7)cm`
`=1.03xx10^(-5) cm=1.03xx10^(-7)m`
`=103xx10^(-9)m=103 nm`.