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Question
- A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state.
- Show that the total number of lines in emission spectrum is `("n"("n - 1"))/2`.
Compute the total number of possible lines in emission spectrum as given in(a).
Solution
- Wavelength, λ = 97.5 nm = 97.5 x 10-9 m
Principle quantum number n = ?
According to Bohr atom model,
`1/lambda = "R" (1/"n"_1^2 - 1/"n"_2^2)` ....n1 = 1; n2 = 1
`1/(lambda"R") = 1 - 1/"n"^2`
`therefore "n" = sqrt((lambda"R")/(lambda"R" - 1))`
Rydberg constant, R = `1.09737 xx 10^7 "m"^-1`
n = `sqrt((97.5 xx 10^-9 xx 1.09737 xx 10^7)/(97.5 xx 10^-9 xx 1.09737 xx 10^7) - 1) = sqrt(1.07/0.07)`
n = 4 - A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level
So total number of lines in emission spectrum is `("n"("n - 1"))/2 = (4(4 - 1))/2 = (4 xx 3)/2` = 6
So the total number of possible lines in emission spectrum is 6.
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