The wavelength of spectral line emitted by hydrogen atom in the Lyman series is `(16)/(15R)` cm.
What is the value of `n_2`?(R-Rydberg constant)

Options

A. 2
B. 3
C. 4
D. 1

Answer

c

Explanation

wavelength ,`lambda=16/(15R)` (Given)
Leyman series equation is given by,
`1/lambda=R(2/(n_1^(2))-1/(n_2^(2)))`
Where,`n_2^(2)=1`
`1/lambda=R(1/1^(2)-1/(n_1^(2)))`
`implies 15R/16=R(1/1^(2)-1/(n_3^(2)))=R((n_2^(2)-1)/(n_2^(2)))`
1`5((n_2^(2))=16((n_2^(2)-1)=16n_2^(2)-16`
`n_2^(2)=16`
`therefore n_2=4`