Answer to Question #88517 in Quantum Mechanics for Doubra

Question

Calculate in ev the energy of a quantum of light wavelength 53000A

Answer 1

We can find the energy of a quantum of light from the formula:

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

here, $E$ is the energy of a quantum of light, $h = 4.135 \cdot 10^{-15} eV \cdot s$ is Plank's constant, $c = 3 \cdot 10^8 m/s$ is the speed of light, $\lambda = 53000 \text{\AA}$ is the wavelength of the light.

Then, we get:

$$E = \dfrac{hc}{\lambda} = \dfrac{4.135 \cdot 10^{-15} eV \cdot s \cdot 3 \cdot 10^8 \dfrac{m}{s}}{53000 \text{\AA} \cdot \dfrac{1 m}{10^{10} \text{\AA} }} = 0.234 eV.$$

Answer:

$E = 0.234 eV.$