Answer to Question #92298 in General Chemistry for tam

Question

Calculate the energy of light emitted as an electron relaxes from the n = 5 to the n = 3 energy level of a hydrogen atom.

Answer 1

The energy of light emitted can be calculated as a difference between energy of the upper level and the lower level:

$$E=E_{n2}-E_{n1}=-\frac{me^4}{8h^2e_0^2}\frac{1}{n_2^2}+\frac{me^4}{8h^2e_0^2}\frac{1}{n_1^2}=\frac{me^4}{8h^2e_0^2}(\frac{1}{n_1^2}-\frac{1}{n_2^2})$$

In our case n₁ = 3 and n₂ = 5:

$$E(light)=\frac{(9.1*10^{-31} kg) (1.6*10^{-19} A*s)^4}{8*(6.63*10^{-34} J*s)^2(8.85*10^{-12} F*m^{-1})^2}(\frac{1}{9}-\frac{1}{25})=1.5*10^{-19} J$$

Thus, the energy of light emitted is equal to 1.5*10^-19 J or 0.94 eV.