The ratio of wavelength of the lest line of Balmer series and the last line Lyman series is:

Options

A. `2`
B. `1`
C. `4`
D. `0.5`

Answer

C

Explanation

Using the relation
`1/lambda=R(1/(n_(1)^(2))-1/(n_(2)^(2)))`
For last line of Blamer series, `n_(1) = 2, n_(2) = oo`
`(1)/(lambda_(B)) = R((1)/(2^(2))-(1)/(oo))=R/4.....(i)`
for last line of Balmer series, `n_(1)=2, n_(2)=oo`
`1/(lambda_(L))=R(1/(1^(2))-1/(oo))=R.........(ii)`
Dividing (ii) by (i), we get
`(lambda_(B))/(lambda_(L))=4`