Calculate the wave number for the longest wavelength transmission in barmer series of atomic hudrogen
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The shortest wavelength transition in the Barlmer series corresponds to the transition
n = 2 → n = ∞. Hence, n1 = 2, n2 = ∞ Balmer.
Answered by
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Answer:
The shortest wavelength transition in the Barlmer series corresponds to the transition
n = 2 → n = ∞. Hence, n1 = 2, n2 = ∞ Balmer.
:
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