Hydrogen atom in its ground state is excited by a radiation of wavelength 97.26 nm. The longest wavelength it emits is

Options

A. 1250 nm
B. 1400 nm
C. 1875 nm
D. 2050 nm

Answer

C

Explanation

`Delta E =(hc)/(lamda) =((6.626xx10^(-34) Js) (3xx10^(8) ms^(-1)))/((97.26xx10^(-9)m))=2.0438xx10^(-18) J`
Also `Delta E=(2.18xx10^(-18) J) ((1)/(1^(2))-(1)/(n_(2)^(2)))`
Hence `(1)/(n_(2)^(2))=1-((Delta E)/(2.18xx10^(-18) J)) =1-(2.0436xx10^(-18))/(2.18xx10%^(-18))=1-0.9374`
`n_(2) = sqrt(1//(1-0.9374)) ~~4`
The transition `n_(2) =4 rarr n_(1)=3` will the longest wavelength. Hence
`Delta E =(2.18xx10^(-18) J) ((1)/(3^(2))-(1)/(4^(2)))=1.06xx10^(-19) J`
`lamda =(hc)/(Delta E) =((6.626xx10^(-34) Js) (3xx10^(8) ms^(-1)))/((1.06xx10^(-19)J))=1.875xx10^(-6) m =1875 nm`