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Question

The frequency of one of the lines in Paschen series of hydrogen atom is 2.750 ×1014 Hz. The quantum number which produces this transition is:


A

n=2

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B

n=4

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C

l=2

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D

n=6

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Solution

The correct option is D

n=6


For paschen series, n1=3, we need to find n2

v=cλ = c×RH[1n21 1n22]

=1λ = c×RH[132 1n2x] = n2 = 6

Solve this using fractions, DO NOT substitute until the end, makes life easier that way.

The thing with the hydrogen atom is that since there is only one electron, all the orbitals - s, p, d, f would be equivalent. Hence when we are at n = 6, l can take any value, and hence l = 2 is correct as well.


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