Calculate wave number for the longest wavelength transition in the Balmer
Series of hydrogen atom.
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v = 1.097× 10^7 [ 1/2^2 - 1/3^2]
v = 1.097× 10^7 [ 1/4 - 1/9]
v = 1.097× 10^7 [ 5/36]
ṽ= 1.5236 × 106 m–1
Answered by
2
Answer:
v = 1.097× 10^7 [ 1/2^2 - 1/3^2]
v = 1.097× 10^7 [ 1/4 - 1/9]
v = 1.097× 10^7 [ 5/36]
ṽ= 1.5236 × 106 m–1
Explanation:
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