Question

class 11 chem Answer it

What electronic transition in the He+ ion would emit the radiation of the same wavelengths as that of the first line Lyman series of hydrogen?
a. n=4 to n=2
v. n=3 to n=2
c. n=5 to n =2
d n=4 to n= 3

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Answer 1

The Rhydberg equation of hydrogen and hydrogen like atoms is :

$:\implies ~\bar{v}= RZ^2~ \bigg\lgroup \dfrac{1}{{n_1}^2}-\dfrac{1}{{n_2}^2}\bigg\rgroup$

For first line of Lyman series of hydrogen atom,

$:\implies~Z=1$

$:\implies~ n_1=1$

$:\implies~ n_2=2$

The wave number is given by :

$:\implies~ \bar{v}= R(1)^2\bigg\lgroup\dfrac{1}{(1)^2}-\dfrac{1}{(2)^2}\bigg\rgroup$

$:\implies~ \bar{v}= R\bigg\lgroup\dfrac{3-1}{4}\bigg\rgroup$

$:\implies~ \bar{v}=\dfrac{3R}{4}$

The wave number is same for the electron transition in $He^+$ ions.

$:\implies~ \bar{v}=RZ^2\bigg\lgroup\dfrac{1}{{n_1}^2}-\dfrac{1}{{n_2}^2}\bigg\rgroup$

$:\implies~ \bigg\lgroup\dfrac{3R}{4}\bigg\rgroup=R(2)^2\bigg\lgroup\dfrac{1}{{n_1}^2}-\dfrac{1}{{n_2}^2}\bigg\rgroup$

$:\implies~ \bigg\lgroup\dfrac{3R}{4}\bigg\rgroup=4R\bigg\lgroup\dfrac{1}{{n_1}^2}-\dfrac{1}{{n_2}^2}\bigg\rgroup$

$:\implies~3R=16R\bigg\lgroup\dfrac{1}{{n_1}^2}-\dfrac{1}{{n_2}^2}\bigg\rgroup$

$:\implies~ \bigg\lgroup\dfrac{3}{16}\bigg\rgroup=\bigg\lgroup\dfrac{1}{{n_1}^2}-\dfrac{1}{{n_2}^2}\bigg\rgroup$

By trial and error method, we get the values :

$:\implies~n_1=2$

$:\implies~ n_2=4$

The electron is undergoing transition from $n=4$ to $n=2.$

Answer : OptionA ( n = 4 to n = 2 )