Questions
If the wavelength of first member of Balmer series of hydrogen spectrum is 6564 , the wavelength of second member of Balmer series will be :
detailed solution
Correct option is B
1λα1n12−1n22∴λ1λ2=122−142122−132⇒6564λ2=2720;∴λ2=4862A∘Talk to our academic expert!
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