Calculate the energy of radiation emitted for the electronic transition from infinity to ground state for hydrogen atom. Given `c = 3xx10^(8)" ms"^(-1), R_(H) = 1.09678 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34)" J s"^(-1)`

Options

A. `2.18xx10^(-18) J`
B. `3.25xx10^(-18) J`
C. `4.05xx10^(-18) J`
D. `2.39xx10^(-18) J`

Answer

A

Explanation

According to Rydberg equation, we have
`bar v=(1)/(lamda) =RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]` where Z=1, `R=1.09678xx10^(7) m^(-1), n_(1)=1 and n_(2)=oo`
Therefore, `(1)/(lamda) =1.09678xx10^(7) [(1)/(1)-(1)/(oo)]" or "lamda =(1)/(1.09678xx10^(7))=0.911xx10^(-7) =9.11xx10^(-8) m`
The energy is given by `E=(hc)/(lamda) =(6.6256xx10^(-34)xx3xx10^(8))/(9.11xx10^(-8))=2.18xx10^(-18) J`