Question

1. A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state.
2. 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).

Answer 1

1. 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
2. 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.