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

Wavelength of spectral lines are given vby
`(1)/(gamma)=(Z^(2)R(1))/(n_(1)^(2))-(1)/(n_(2)^(2))`
For last line of Balmer series ,
`n_(1)=2 and n_(2) = oorArr(1)/(gamma_(B))=Z^(2)R(1)/(2^(2))-(1)/(oo^(2))=R/4[beforeZ=1]`
Similarly for last line of Lyman series.
`n_(1)=1and n_(2) = oo`
`rArr (1)/(lambda_(2))=Z^(2)R(1)/(1^(2))-(1)/(oo^(2))=RrArr((1)/(gamm_(B)/(1)/(gamma_(L))=(R)/(4)/(R)=(1)/(4)`
`rArr (gamma_(L))/(gamma_(B))=(1)/(4)rArr(gamma_(B))/(gamma_(L))=4`