11. An electron jumps from 3rd orbit to 1st orbit in H-atom. Find out the wavelength of emitted radiations?

1 Answer
Jun 24, 2018

lambda = 102.6 color(white)(l) nm

Explanation:

The energy of the photon emitted shall equal to the difference between the electron's potential energy at the two orbits.

The ul("Rydberg Formula") suggests that an electron that travels (or literally "falls") from energy level n_i to n_f (n_i>n_f) emits a photon of wavelength lambda for which

1/lambda=R(1/n_f^2-1/n_i^2)

where the Rydberg's Constant R=1.097 xx 10^7 color(white)(l) m^{-1}.

In this particular scenario

  • n_i=3
  • n_f=1

Such that

1/lambda = R(1/n_f^2-1/n_i^2)
color(white)(Delta "PE") = 1.097 xx 10^7 color(white)(l) m^(-1)* (1/1^2-1/3^2)
color(white)(Delta "PE") = 9.751*10^7 color(white)(l) m^(-1)

lambda = 1.026 xx 10^(-7) color(white)(l) color(red)(cancel(color(black)(m))) * (10^9 color(white)(l) nm)/(1 color(white)(l) color(red)(cancel(color(black)(m))))
color(white)(lambda) = 102.6 color(white)(l) nm

References
"Bohr's Hydrogen Atom" , Chemistry Libretext