Question

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

Answer 1

`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