What will be the longest wavelength line in balmer series of spectrum of h-atom?
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163
Since wavelength and energy are inversely related, the longest wavelength would be produced by the lowest amount of energy. That would be by the electron falling from from n = 3 (ni) to n = 2 (nf). (All Balmer series lines have nf = 2).
Using the Rydberg equation,
1 / lambda = R (1/nf^2 - 1/ni^2) = (1.097 x 10^7 m^-1)(1/(2^2) - 1/(3^2)) = 1.52 x 10^6 m^-1
lambda = 1 / (1.52 x 10^6 m) = 6.56 x 10^-7 m = 656 nm . . .that's a RED line.
Hope my answer is helpful to u.
Using the Rydberg equation,
1 / lambda = R (1/nf^2 - 1/ni^2) = (1.097 x 10^7 m^-1)(1/(2^2) - 1/(3^2)) = 1.52 x 10^6 m^-1
lambda = 1 / (1.52 x 10^6 m) = 6.56 x 10^-7 m = 656 nm . . .that's a RED line.
Hope my answer is helpful to u.
Answered by
7
Answer:
656nm will be the answer
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