What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
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1 Answer
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For an atom? = 1/ λ = RH Z2 [ (1/n12) – (1/n22)]
For He+ spectrum: Z = 4, n2 = 4, n1 = 2.
∴? = 1/ λ = RH Z2 [ (1/22) – (1/42)]
= RH 22 [ (1/22) – (1/42)]
= 3RH /4
For hydrogen spectrum:
∴? = 1/ λ = RH 12 [ (1/n12) – (1/n22)] = 3RH /4
=> (1/n12) – (1/n22) = 3/4
This corresponds to n1 = 1 and n2 =2 and means that the transition has taken place in the Lyman series from n = 2 to n =1.
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