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Chapter 6: Problem 61

A photon has energy 20 keV. What are its frequency and wavelength?

Short Answer

Expert verified
The energy of the photon is given as 20 keV, which is equal to \(3.20436\times10^{-15} \, \text{J}\) after converting to Joules. Using Planck's equation with \(E = hf\), we find the frequency: \(f = \frac{E}{h} = \frac{3.20436\times10^{-15}\, \text{J}}{6.626\times10^{-34}\, \text{Js}} \approx 4.836\times10^{18}\, \text{Hz}\). Then using the relationship between frequency and wavelength, \(c = \lambda f\), we get: \(\lambda = \frac{c}{f} = \frac{3\times10^8 \, \text{m/s}}{4.836\times10^{18}\, \text{Hz}} \approx 6.203\times10^{-11}\, \text{m}\). So, the frequency is approximately \(4.836\times10^{18} \, \text{Hz}\) and the wavelength is approximately \(6.203\times10^{-11} \, \text{m}\).

Step by step solution

01

Convert energy from keV to Joules

Given the energy of the photon is 20 keV, let's first convert it to Joules. We know that 1 keV = 1.60218 × 10^{-16} J. Therefore, \(E = 20 \,\text{keV} * 1.60218\times10^{-16}\, \frac{\text{J}}{\text{keV}}\)
02

Find the frequency using Planck's equation

Planck's equation relates the energy of a photon (E) to its frequency (f) as: \(E = hf\) where h is the Planck's constant, which equals \(6.626 \times 10^{-34} \text{Js}\). We can calculate the frequency (f) by rearranging the equation: \(f = \frac{E}{h}\) Substitute the energy value from Step 1 and the Planck's constant into the equation, and solve for f.
03

Find the wavelength using the frequency-wavelength relation

The relationship between frequency (f) and wavelength (λ) is given by the formula: \(c = \lambda f\) where c is the speed of light, approximately \(3\times10^8 \, \text{m/s}\). We can calculate the wavelength (λ) by rearranging the equation: \(\lambda = \frac{c}{f}\) Substitute the frequency value from Step 2 and the speed of light into the equation, and solve for λ.

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