Chapter 27, Problem 67GP

To determine

The number of microwave photons should be determined.

Answer to Problem 67GP

  $\text{4}{\text{.66×10}}^{\text{26}}\text{photons/s}$

Explanation of Solution

Given info: Wavelength, $\text{λ = 12}\text{.2}\text{cm}\text{= 12}\text{.2}{\text{×10}}^{\text{-2}}\text{m}$

Power, $\text{P}\text{=}\text{760}\text{W}$ .

Formula used:

The energy of the photon is

  $\text{E}\text{=}\text{hf}\text{=}\frac{\text{hc}}{\text{λ}}$

  $f$ is frequency.

  $p$ is photon’s momentum

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

Calculation:

Substituting the given values in the formula, $\text{E}\text{=}\text{hf}\text{=}\frac{\text{hc}}{\text{λ}}$

Where, $c$ is speed of light, ${\text{3×10}}^{\text{8}}\text{ m/s }$

  $\text{E}\text{=}\frac{\left(\text{6}\text{.63}{\text{×10}}^{\text{-34}}\text{J}\cdot \text{s}\right)\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\left(\text{12}\text{.2}{\text{×10}}^{\text{-2}}\text{m}\right)}\text{=1}\text{.63}\text{×}{\text{10}}^{\text{-24}}\text{J}$

Now calculate the rate at which photons are produced in the oven:

  $\text{N = }\frac{\text{P}}{\text{E}}\text{ = }\frac{\left(\text{760}\text{W}\right)}{\text{1}{\text{.63×10}}^{\text{-24}}\text{J}}\text{= 4}{\text{.66×10}}^{\text{26}}\text{photons/s}$

Conclusion:

The required value is $\text{4}{\text{.66×10}}^{\text{26}}\text{photons/s}$ .