Question

calculate the wavelength at which the energy of a photon becomes equal to the average thermal energy of atoms in a solid at room temperature.

Answer 1

Answer on Question 49878, Physics, Optics

Question:

Calculate the wavelength at which the energy of a photon becomes equal to the average thermal energy of atoms in a solid at room temperature.

Solution:

From the definition of the energy of the photon we have:

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

where $E$ is the energy of the photon, $h$ is the Planck's constant, $c$ is the speed of the light and $\lambda$ is the wavelength of the light.

The average thermal energy of atoms in a solid is $E = k_{B}T$, where $k_{B}$ is Boltzmann constant, $T$ is the temperature.

According to the condition of the question we equate both relationships for energy:

$$\frac{hc}{\lambda} = k_{B}T.$$

From this equation we obtain the wavelength at which the energy of a photon becomes equal to the average thermal energy of atoms in a solid at room temperature:

$$\lambda = \frac{hc}{k_{B}T} = \frac{6.626 \cdot 10^{-34} \, \text{Js} \cdot 3 \cdot 10^{8} \, \frac{\text{m}}{\text{s}}}{1.38 \cdot 10^{-23} \, \frac{\text{J}}{\text{K}} \cdot 298.15 \, \text{K}} = 4.8 \cdot 10^{-5} \, \text{m} = 48 \, \mu\text{m}.$$

Answer:

$$\lambda = 48 \, \mu\text{m}.$$

http://www.AssignmentExpert.com/