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Q.

The longest wave length of hydrogen atom in Lyman series is λ. The shortest wavelength in Balmer series of He+ is

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a

5λ9

b

34λ

c

3λ4

d

59λ

answer is C.

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Detailed Solution

For H:-  1λH=RH[112122]×12
For  He+:1λHe+=RH[1221α2]×22
1λH=3RH4  …… (1)
1λHe+=RH[14]×4=RH  ….. (2)
From (1) & (2) λHe+λ=1RH.3RH4
λHe+=3λ4

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