Hydrogen `(._(1)H^(1))`, Deuterum `(._(1)H^(2))`, singly ionised Hellium `(._(2)He^(4))^(+)` and doubly ionised lithium `(._(3)Li^(6))^(++)` all have one electron around the nucleus. Consider an electron tranition from `n=2` to `n=1`. If the wave lengths of emitted radiation are `lambda_(1),lambda_(2),lambda_(3)` and `lambda_(4)` respectively then approximately which one of the follwing is correct ?

Options

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

Answer

C

Explanation

The wave number of a spectral line is given by,
`1/lambda=RZ^(2)(1/n_(1)^(2)-1/n_(2)^(2))`
For the transition n = 2 to n = 1,
`" "1/lambda=RZ^(2)(1/1^(2)-1/2^(2))=3/4RZ^(2) therefore lambda=4/(3RZ^(2))`
For `H_(2), Z=1, therefore lambda_(1)=4/(3R)" ...(i)"`
For deuterium, `Z=1, therefore lambda_(2)=4/(3R)" ...(ii)"`
For singly ionized helium, Z = 2
`therefore lambda_(3)=4/(3R(4))=4/(12R) therefore 4lambda_(3)=4/(3R)" ...(iii)"`
and for lithium, Z = 3 and `lambda_(4)=4/(3R(9))=4/(27R)`
`therefore 9lambda_(4)=4/(3R)" ...(iv)"`
Thus, from (i), (ii), (iii) and (iv) we get
`lambda_(1)=lambda_(2)=4lambda_(3)=9lambda_(4)`