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

Answer

B

Explanation

` (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)`
For hydrogen atom
`(1)/(91.2) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)`
For `He^(Theta)`
` (1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))Z^(2)`
Dividing the equation we get
`(lambda)/(91.2) = (1)/(4)`
`lambda = 22.8 nm`