Question

problem 27.31
part c
the quantity \\( \lambda_{\text {com }} \\), which has the dimensions of length, is called the compton wavelength.
select the correct equations that show that if a photon has wavelength equal to the compton wavelength of a particle with mass \\( m \\), the photons energy is equal to the rest energy of the particle, \\( m c^{2} \\).
\\( \bigcirc \\) \\( e_{\text {photon }} = h f = \frac{h c}{\lambda} = \frac{h c}{\lambda_{\text {com }}} = \frac{h c}{h /(m c)} = m c^{2} \\)
\\( \bigcirc \\) \\( e_{\text {photon }} = \frac{h}{f} = h \lambda c = h (h /(m c)) c = m c^{2} \\)
\\( \bigcirc \\) \\( e_{\text {photon }} = h f = \frac{h c}{\lambda} = \frac{h c}{h /(m c)} = m c^{2} \\)
\\( \bigcirc \\) \\( e_{\text {photon }} = \frac{h}{f} = \frac{h \lambda}{c} = \frac{h (m c / h)}{c} = m c^{2} \\)
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incorrect; try again; 2 attempts remaining

Explanation:

Step1: Recall Compton Wavelength Formula

The Compton wavelength of a particle is given by $\lambda_c = \frac{h}{mc}$, where $h$ is Planck's constant, $m$ is the mass of the particle, and $c$ is the speed of light.

Step2: Recall Photon Energy Formula

The energy of a photon is given by $E = hf$, and since $c = \lambda f$ (where $\lambda$ is the wavelength of the photon), we can also write $E=\frac{hc}{\lambda}$.

Step3: Substitute Compton Wavelength into Photon Energy

If the photon's wavelength $\lambda$ is equal to the Compton wavelength $\lambda_c=\frac{h}{mc}$, then substitute $\lambda = \frac{h}{mc}$ into the photon energy formula:

$$ E_{\text{photon}}=\frac{hc}{\lambda}=\frac{hc}{\frac{h}{mc}} = mc^2 $$

Now let's check the options:

• Option 1: The steps here have an error in the substitution (incorrect manipulation of $c$ and $\lambda$).
• Option 2: Incorrect formula for photon energy (wrongly uses $h/\lambda$ and incorrect manipulations).
• Option 3: $E_{\text{photon}} = hf=\frac{hc}{\lambda}=\frac{hc}{\frac{h}{mc}}=mc^2$ (this follows the correct substitution of Compton wavelength into photon energy formula).
• Option 4: Incorrect formula for photon energy and wrong manipulations.

Answer:

The correct equation is $E_{\text{photon}} = hf=\frac{hc}{\lambda}=\frac{hc}{\frac{h}{mc}}=mc^2$ (the third option in the list).