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Question

The approximate wave number of the spectral line of the shortest wavelength in Balmer series of atomic hydrogen will be:
(Rydberg constant (RH)=109678 cm1)

A
22345.8 cm1
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B
37515.6 cm1
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C
15519.8 cm1
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D
27419.5 cm1
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Solution

The correct option is D 27419.5 cm1
Shortest wavelength means maximum energy.
Therefore, the electronic transition involved should be
n2=n1=2 (as it belongs to the Balmer series)
We know,
1λ=RZ2(1n211n22)
Where, λ=Wavelength; R=Rydberg constant
1λshortest=ν=109678×12×(12212)
ν=27419.5 cm1

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