Wavelength of high energy transition of hydrogen atoms is 91.2 nm. Calculate the corresponding wavelength of He atoms.

Options

A. 22.8 nm
B. 20.5 nm
C. 14.6 nm
D. 17.9 nm

Answer

A

Explanation

For maximum energy, `n_(1)=1 and n_(2)=oo`. Hence, we have
`(1)/(lamda) prop Z^(2) ((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))` Since `R_(H)` is constant and transition remains the same , `(1)/(lamda) prop Z^(2) rArr (lamda_(He))/(lamda_(H))=(Z_(H)^(2))/(Z_(He)^(2))=`
Hence, `lamda_(He) =(1)/(4) xx91.2=22.8 nm`