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Question

Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.
Rydberg constant = 109677 cm1

A
109677.84 cm1
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B
15232.92 cm1
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C
27419.25 cm1
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D
20564.44 cm1
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Solution

The correct option is C 27419.25 cm1
The shortest wavelength transition in the Balmer series corresponds to the transition: n=2n=
The shortest wavelength requires the highest energy for an electron to jump from n = 2 to n=
So, n1=2 and n2=
Using the formula:
¯v=RH(1n211n22) where RH = Rydberg constant
¯v=(109677 cm1)(12212)On solving,¯v=27419.25 cm1

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