Question

The wavelength '$\lambda$' of a photon and the deBroglie wavelength of an electron have same value. The ratio of kinetic energy of the electron to the energy of a photon is ______.

• A. $\frac{2\lambda mc}{h}$
• B. $\frac{\lambda mc}{h}$
• C. $\frac{h}{2\lambda mc}$
• D. $\frac{h}{\lambda mc}$

Hint:

Whenever an electron and a photon have the exact same wavelength (and thus the same momentum), the photon will ALWAYS have significantly more energy than the electron because $c \gg v$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must find the mathematical ratio between the kinetic energy of an electron and the total energy of a photon, given that their wavelengths are exactly identical.

Step 2: Detailed Explanation:
1. Energy of the Photon ($E_p$):
The energy of a photon with wavelength $\lambda$ is given by the standard Planck-Einstein relation:
$E_p = \frac{hc}{\lambda}$
2. Kinetic Energy of the Electron ($K_e$):
The de Broglie wavelength of an electron moving with momentum $p$ is $\lambda = \frac{h}{p}$.
Rearranging for momentum gives $p = \frac{h}{\lambda}$.
The kinetic energy of a non-relativistic particle in terms of its momentum is $K_e = \frac{p^2}{2m}$.
Substitute the momentum expression into the kinetic energy formula:
$K_e = \frac{\left(\frac{h}{\lambda}\right)^2}{2m} = \frac{h^2}{2m\lambda^2}$
3. Calculate the Ratio:
We need the ratio of the electron's kinetic energy to the photon's energy:
$\text{Ratio} = \frac{K_e}{E_p}$
$\text{Ratio} = \frac{ \frac{h^2}{2m\lambda^2} }{ \frac{hc}{\lambda} }$
Multiply by the reciprocal of the denominator:
$\text{Ratio} = \frac{h^2}{2m\lambda^2} \times \frac{\lambda}{hc}$
Cancel out one '$h$' from the numerator and denominator, and one '$\lambda$' from the numerator and denominator:
$\text{Ratio} = \frac{h}{2mc\lambda}$

Step 3: Final Answer:
The ratio is $\frac{h}{2\lambda mc}$, matching option (c).