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Step 1: State the de Broglie wavelength condition
An electron moving at speed $v$ and a photon moving at speed $c$ are given to have the same de Broglie wavelength. By definition of de Broglie wavelength, we have:
$$
\lambda = \frac{h}{p},
$$
where $h$ is Planck’s constant and $p$ is the momentum of the particle. Since both have the same de Broglie wavelength:
$$
\frac{h}{p_e} = \frac{h}{p_{ph}} \quad \text{or} \quad p_e = p_{ph}.
$$
Step 2: Write the momenta of electron and photon
• The momentum of the electron is
$$
p_e = m \, v,
$$
where $m$ is the mass of the electron and $v$ is its speed.
• The momentum of the photon is given by
$$
p_{ph} = \frac{E_{ph}}{c},
$$
where $E_{ph}$ is the energy of the photon and $c$ is the speed of light.
From $p_e = p_{ph}$, we get:
$$
m \, v = \frac{E_{ph}}{c}.
$$
Step 3: Express the kinetic energy of the electron and photon
• The kinetic energy of the electron, $k_e$, is related to its momentum by:
$$
k_e = \frac{p_e^2}{2m} = \frac{(m \, v)^2}{2m} = \frac{m\,v^2}{2}.
$$
• The energy of a photon is
$$
E_{ph} = p_{ph} \, c.
$$
Step 4: Relate electron's kinetic energy to photon's energy
From $m \, v = \frac{E_{ph}}{c}$, we can find $v$:
$$
v = \frac{E_{ph}}{m \, c}.
$$
Hence, the kinetic energy of the electron becomes:
$$
k_e = \frac{1}{2} m \left(\frac{E_{ph}}{m \, c}\right)^2
= \frac{1}{2} m \frac{E_{ph}^2}{m^2 \, c^2}
= \frac{E_{ph}^2}{2 m \, c^2}.
$$
Step 5: Form the ratio of kinetic energies
Since the photon's energy is $E_{ph}$, the ratio of the electron's kinetic energy to the photon's energy is:
$$
\frac{k_e}{E_{ph}}
= \frac{\tfrac{E_{ph}^2}{2 m \, c^2}}{E_{ph}}
= \frac{E_{ph}}{2 m \, c^2}.
$$
Using $E_{ph} = p_{ph} \, c = p_e \, c = m \, v \, c$, we substitute $E_{ph} = m \, v \, c$ into the above expression:
$$
\frac{k_e}{E_{ph}}
= \frac{m \, v \, c}{2 m \, c^2}
= \frac{v}{2c}.
$$
Therefore,
$$
\frac{k_e}{E_{ph}} = \frac{v}{2c}.
$$
Answer: The ratio of the electron's kinetic energy to the photon's energy is
$$
\frac{v}{2c}.
$$
