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Q.

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is …

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a

2430 Å

b

1215 Å

c

4687 Å

d

1640 Å

answer is A.

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Detailed Solution

First line of Balmer of Hydrogen
$\frac{1}{λ_{1}} = R {(1)}^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}] where λ_{1} =6561$
Second line of Balmer of singly ionized Helium
$\frac{1}{λ_{2}} = R {(2)}^{2} [\frac{1}{2^{2}} - \frac{1}{4^{2}}]$
Dividing $λ_{2} = λ_{1} x \frac{5}{27} = 6561 x \frac{5}{27} = 1215 A^{0}$

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