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Q.

The wavelength of the energy emitted when electron come from fourth orbit to second orbit in hydrogen is 20.397 cm.  The wavelength of energy for the same transition in He+  is

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a

5.099 cm-1

b

20.497 cm-1

c

40.994 cm-1

d

81.988 cm-1

answer is A.

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Detailed Solution

E=hcλ∝Z2n2⇒λ∝1Z2 Hence λHe+=20.3974=5.099 cm
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