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 Concept:
For a photon, energy is $E_p = \frac{hc}{\lambda}$. For an electron, the deBroglie wavelength is $\lambda = \frac{h}{\sqrt{2mK_e}}$, where $K_e$ is the kinetic energy.

Step 2: Formula Application:
From the deBroglie relation for the electron: $\lambda^2 = \frac{h^2}{2mK_e} \implies K_e = \frac{h^2}{2m\lambda^2}$.

Step 3: Explanation:
The ratio $\frac{K_e}{E_p} = \frac{h^2 / (2m\lambda^2)}{hc / \lambda}$. Simplifying the fraction: $\frac{K_e}{E_p} = \frac{h^2}{2m\lambda^2} \times \frac{\lambda}{hc} = \frac{h}{2m\lambda c}$.

Step 4: Final Answer:
The ratio is $\frac{h}{2\lambda mc}$.