Question

calculate the shortest wavelength of electron which is present in Balmer's series​

Answer 1

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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Answer 2

The answer is 364.6 nm.....

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