Answer to Question #156373 in Physics for SallyJane

Question

The bright light emitted by sodium lamps comes from a transition between electrons at an energy level of – 3.04 eV falling to an energy level of -5.14 eV. Calculate the wavelength of the light emitted

Answer 1

The wavelenght $\lambda$ of the light is connected with its energy $E$ in the following way:

$$E = \dfrac{hc}{\lambda}$$

where $h = 6.58\times 10^{-16}eV\cdot s$ is the Plank's constant, and $c = 3\times 10^{8} m/s$ is the speed of light.

On the other hand, the energy of the emitted light is equal to the difference between electon's energy levels:

$$E = |-5.14eV - 3.04eV| = 8.18eV$$

Expressing $\lambda$, obtain:

$$\lambda = \dfrac{hc}{E}\\ \lambda = \dfrac{6.58\times 10^{-16}\cdot 3\times 10^8}{8.18} \approx 2.41\times 10^{-8}m$$

Answer. $2.41\times 10^{-8}m$.