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Question

Find out the electronic state from which an electron drops to emit radiation with a wavelength of 926oA in ultraviolet region (lyman series) of spectrum of hydrogen atom.
Rydberg constant = 1.097×107m1

A
n = 4
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B
n = 5
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C
n = 6
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D
n = 8
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Solution

The correct option is D n = 8
Given: wavelength of the radiation = λ=926A=926×1010m
We know,
¯v=1λ=RH(1n211n22)
where, ¯v = wave number of the radiation.
Since the radiation is in the U.V region and belongs to Lyman series therefore n1=1
¯v=1926×1010m=1.097×107m1(1121n22)11n22=19.26×1010×1.097×107
On solving, n2 = 8

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