Hydrogen `(_(1)H^(1))` Deuterium `(_(1)H^(2))` singly omised helium `(_(1)He^(1))` and doubly ionised lithium `(_(1)Li^(6))^(++)` all have one electron around the nucleus Consider an electron transition from `n = 2 to n = 1` if the wavelength of emitted radiartion are `lambda_(1),lambda_(2), lambda_(3),and lambda_(4)`, repectivelly then approximetely which one of the following is correct ?

Options

A. `lamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)`
B. `lamda_(1)=2lamda_(2)=3lamda_(3)=4lamda_(4)`
C. `4lamda_(1)=2lamda_(2)=2lamda_(3)=lamda_(4)`
D. `lamda_(1)=2lamda_(2)=2lamda_(3)=lamda_(4)`

Answer

A

Explanation

We know `(1)/(lamda)=RZ^(2)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=RZ^(2)[(1)/(1^(2))-(1)/(2^(2))]`
`lamda=(4)/(3RZ^(2))`
`lamda_(1)=(4)/(3R),lamda=(4)/(3R)k,lamda_(3)=(4)/(12R),lamda_(4)=(4)/(27R)`
`thereforelamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)`.