Q. During the emission spectrum,the first line of Lyman series of H-atom occurs as `lambda` = ` ``x ` .The wavelength of the 1st line of Lyman series of `He^+` ion will occur at:-

A) `(4/x)`

B) `3x`

C) `(3/x)`

D) `4x`

E) `(x/4)`

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The Rydberg equation for transition of an electron from n=2 to n=1 in hydrogen-like elements can be expressed as:

`1/lambda=R_oo*Z^2(1/n_1^2-1/n_2^2)`

Where, lambda is the wavelength of the emitted radiation, R_oo is the Rydberg constant and Z is the atomic number of the element.

For hydrogen, Z=1, for the first line of transition of Lyman series, `n_2=2, n_1=1,`

So, `1/x=R_oo *1^2(1/ 1^2-1/ 2^2)`

`rArr x=4/(3R_oo)`

`rArr R_oo=4/(3x)`

Putting this value of `R_oo` in `He^(2+)(Z=2)` , for the first line of transition,

`1/lambda=4/(3x)*2^2(1/ 1^2-1/ 2^2)`

`rArr 1/lambda=4/(3x)*4*3/4=4/x`

`rArr lambda=x/4`

Hence the correct answer is option E).

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