The ionization energy of a hydrogen atom...

The ionization energy of a hydrogen atom in terms of Rydberg constant `(R_(H))` is given by the expression

`R_H hc`

`R_H c`

`2R_H hc`

`R_H N_A hc`

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The correct Answer is:

`1/lambda=R_Hz^2[1/n_1^2-1/n_2^2]`
For ionization energy
`n_1=1` and `n_2=oo` `implies 1/lambda=R_Himplies IE_1=E_oo-E_1=(hc)/lambda=R_Hhc`

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