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The wavelength of the first line of Lyma...

The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate the atomic no of ion

A

`1`

B

`3`

C

`4`

D

`2`

Text Solution

Verified by Experts

The correct Answer is:
D
Wavelength of the first line of Lyman series for hydrogen atom will be given by the equation
`(1)/(lambda_1) = R((1)/(1^2) - (1)/(2^2)) = (3R)/(4)`
The wavelength of second Balmaer line for hydrogen like ion x is
(1)/(lambda_2) = RZ^2((1)/(2^2) - (1)/(4^2)) = (3RZ^2)/(16)`
Given that `lambda__1= lambda_2 or (1)/(lambda_1) =(1)/(lambda_2)`
i.e. (3R)/(4) = (3RZ)^2/(16)`
`:. Z=2`
i.e. x ion is `He^+`. The energies of first four levels of x are
`E_1 = - (13.6)Z^2 =- 54.4e V`
`E_2 = (E_1)/(2)^2 =-13.6 eV`
`E_3 = (E_1)/(3)^2 =- 6.04 eV`
and `E_4 = (E_1)/(4)^2 =- 3.4 eV`
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