The first line of the lyman series in a hydrogen spectrum has a wavelength of 1210 Å. The corresponding line of a hydrogen like atom of `Z=11` is equal to

Options

A. `4000Å`
B. `100Å`
C. `40Å`
D. `10Å`

Answer

D

Explanation

By Bohr.s formula `(1)/(lambda)=Z^(2)R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
For first line of Lyman series, `n_(1)=1, n_(2)=2`
`therefore (1)/(lambda)=Z^(2)R(3)/(4)`
In the case of hydrogen atom, Z = 1
`(1)/(lambda)=R(3)/(4)`
For hydrogen - like atom, `Z= 11`
`(1)/(lambda.)=121R(3)/(4)`
`(lambda.)/(lambda)=(3R)/(4)xx(4)/(121Rxx3)=(1)/(121) rArr lambda.=(lambda)/(121)=(1210)/(121)=10Å`