Question

Calculate the wave number for the longest wavelength transmission in barmer series of atomic hudrogen

Answer 1

The shortest wavelength transition in the Barlmer series corresponds to the transition

n = 2 → n = ∞. Hence, n1 = 2, n2 = ∞ Balmer.

Answer 2

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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