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Hydrogen (.(1)H^(1)), Deuterum (.(1)H^(2...

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 ?

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)`

Text Solution

Verified by Experts

The correct Answer is:
C

Since , `1/lambda = RZ^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]`
For , `._(1)H^(1)1/(lambda_(1))= R(1)^(2) (1-1/4)=(3R)/4`
For `._(1)H^(1)1/(lambda_(1))=R(1)^(2)(1-1/4)=(3R)/4`
For , `._(1)H^(2)1/(lambda_(2))=R(1)^(2)(1-1/4)=(3R)/4`
For `(._(3)Li^(6))^(++) 1/(lambda_(4))=R(3)^(2)(1-1/4)=(27R)/4 = 9/(lambda_(1))`
` :. lambda_(1)= lambda_(2)=4lambda_(3) = 9 lambda_(4)`
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