How to calculate the wave number for shortest wavelength transition in the Balmer Series of atomic Hydrogen?
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The shortest wavelength transition in the Barlmer series corresponds to the transitionn = 2 → n = ∞. Hence, n1 = 2, n2 = ∞ Balmer
Putting the values in the equation we get, wavenumber=27419.25/cm.
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- The shortest wavelength transition in the Barlmer series corresponds to the transitionn = 2 → n = ∞. Hence, n1 = 2, n2 = ∞ Balmer
- Putting the values in the equation we get, wavenumber=27419.25/cm.
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