Respected Sir/Madam Please help me with Que. 3 as shown in the image.



This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

1 Answer | Add Yours

Top Answer

jerichorayel's profile pic

Posted on (Answer #1)

When an atom goes back from its excited state to the ground state, it will omit photon of light which is equal to the change between two energy levels. We can manipulate the Rydberg formula for hydrogen atom in order to get the principal quantum number of the excited state.

`1/(lambda) = RZ^(2) (1/n_(1)^(2)-1/(n_(2)^(2)))`

Z = 1 (for hydrogen)

n1 = 1 

`1/(lambda) = R*1^(2) (1/1^(2)-1/(n_(2)^(2)))`

`1/(lambda) = R (1-1/(n_(2)^(2)))`

`1/(lambda* R) = (1-1/(n_(2)^(2)))`

`1/(n_(2)^(2)) = 1 -1/(lambda* R)`

`1/(n_(2)^(2)) = (lambda* R - 1)/(lambda* R)`

`n_(2)^(2) =(lambda* R)/(lambda* R - 1)`

`sqrt(n_(2)^(2) =(lambda* R)/(lambda* R - 1))`

`n_(2) = sqrt((lambda* R)/(lambda* R - 1))`

`n_(2) = [(lambda* R)/(lambda* R - 1)]^(1/2)`


The answer is letter A. 


We’ve answered 288,059 questions. We can answer yours, too.

Ask a question