The wavelength of high energy transition of H atoms is `91.2 nm` Calculate the corresponding wavelength of He atom.

Options

A. 2.28 nm
B. 22.8 nm
C. 182.4 nm
D. 364.8 nm

Answer

B

Explanation

According to Rydberg’s equation
`bar(v) =(1)/(lambda) = RZ^(2) [ (1)/(n_(L)^(2))-(1)/(n_(H)^(2))]`
where
`bar(v)` = wave number
`lambda` = wavelength
R =Rydberg constant (=109677 `cm^(-1))`
Z = atomic number
`n_(L)` = ower energy state
`n_(H)` = higher energy state
i.e `(1)/(lambda) prop Z^(2)`
Thus `(lambda(He^(+)))/(lambda(H)) = (Z^(2)(H))/(Z^(2)(He^(+))`
To find `lambda (He) =?`
Given `lambda(H) = 91.2 `nm
Z(H) =1
Z(He) =2
`:. lambda (He^(+)) = (1xx 91.2)/(4) =22.8` nm