Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` then

Options

A. `(16)/(lambda_(1))=(9)/(lambda_(2))`
B. `(16)/(lambda_(2))=(3)/(lambda_(1))`
C. `(4)/(lambda_(1))=(1)/(lambda_(2))`
D. `(16)/(lambda_(1))=(3)/(lambda_(2))`

Answer

2

Explanation

`(1)/(lambda_(1))=R(1)^(2)[(1)/(1^(2))-(1)/(oo^(2))]" "` and `" "(1)/(lambda_(2))=R(1)^(2)[(1)/(2^(2))-(1)/(4^(2))]`
`:. " "lambda_(1)=(1)/(R)" "`and `" "lambda_(2)=(16)/(3R)" ":." "(16)/(lambda_(2))=(3)/(lambda_(1))`