The first line in the Lyman series has wavelength `lambda`. The wavelegnth of the first line in Balmer series is

Options

A. `(5)/(27)lamda`
B. `(36)/(5)lamda`
C. `(27)/(5)lamda`
D. `(5)/(36)lamda`

Answer

C

Explanation

According to Bohr, the wavelength emitted when an electron jumps from `n_(1)^(th)` to `n_(2)^(th)` orbit is
`E=(hc)/(lamda)=E_(2)-E_(1)`
`=(2pi^(2)mk^(2)e^(4))/(h^(2))[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
or `(1)/(lamda)=(2pi^(2)mk^(2)e^(4))/(ch^(3))((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
where `k=(1)/(4piepsi_(0)) and (2pi^(2)mk^(2)e^(4))/(ch^(3))` is R (Rydberg constant)
`implies (1)/(lamda)=R((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For first line in Lyman series
`(1)/(lamda_(L))=R((1)/(1^(2))-(1)/(2^(2)))=(3R)/(4)` . .. (i)
For first line in Balmer series,
`(1)/(lamda_(B))=R(1(1)/(2^(2))-(1)/(3^(2)))=(5R)/(36)` . .. (ii)
From eqs. (i) and (ii)
`therefore (lamda_(B))/(lamda_(L))=(3R)/(4)xx(36)/(5R)=(27)/(5)`
`therefore lamda_(B)=(27)/(5)lamda" "(because lamda_(L)=lamda)`