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

Answer

D

Explanation

For first line of lyman series `n_1=1`,`n_2=2`
`therefore (1)/(lambda)=Z^2R(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)`
`rArr(lambda')/(lambda)=(3R)/(4)xx(4)/(121Rxx3)=(1)/(121)rArrlambda'=(lambda)/(121)=(1210)/(121)=10Å`