calculate the shortest wavelength of electron which is present in Balmer's series
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calculate the shortest wavelength of electron which is present in Balmer's series
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Answer:
364.6 nm
Explanation:
wavelength =
1/λ = R(1/n1^2 - n2^2)
Shortest wavelength is emitted in Balmer series if the transition of electron takes place from n2 = ∞ to n1 = 2
∴ Shortest wavelength in Balmer series = R(1/2^2 - 1/∞)
Or λ = 4/R
R = rydberg's constant and value = 1.097 * 10^7
substituting the value of R in λ we get,
(4/1.097 * 10^7)
364.6 nm
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The answer is 364.6 nm.....
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