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 = 1.097×107m−1
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∘=926×10−10m We know, ¯v=1λ=RH(1n21−1n22) 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=1926×10−10m=1.097×107m−1(112−1n22)1−1n22=19.26×10−10×1.097×107 On solving, n2 = 8