Question

An electron with speed υ and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E_e and p_e and that of photon are E_ph and p_ph respectively. Which of the following is correct?

• A. \(\frac{E_e}{E_{\text{ph}}} = \frac{2c}{v}\)
• B. \(\frac{E_e}{E_{\text{ph}}} = \frac{v}{2c}\)
• C. \(\frac{p_e}{p_{\text{ph}}} = \frac{2c}{v}\)
• D. \(\frac{p_e}{p_{\text{ph}}} = \frac{v}{2c}\)

Answer 1

The Correct Option is B

To solve the problem, we need to understand the concept of de Broglie wavelength, kinetic energy, and momentum for both electrons and photons.

1. According to de Broglie's hypothesis, the wavelength \(\lambda\) associated with a particle is given by:

\(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

2. For an electron with momentum \(p_e\) and a photon with momentum \(p_{\text{ph}}\), since their de Broglie wavelengths are equal, we can write:

\(\frac{h}{p_e} = \frac{h}{p_{\text{ph}}}\)

This implies:

\(p_e = p_{\text{ph}}\)

3. The kinetic energy \(E_e\) of an electron is given by:

\(E_e = \frac{1}{2}mv^2\)

For non-relativistic speeds, the momentum \(p_e\) is \(mv\), so:

\(p_e = mv\)

4. The energy of a photon \(E_{\text{ph}}\) is given by:

\(E_{\text{ph}} = p_{\text{ph}}c\)

5. Now, substituting \(p_e = p_{\text{ph}}\) in the energy expressions, we compare \(E_e\) and \(E_{\text{ph}}\):

\(\frac{E_e}{E_{\text{ph}}} = \frac{\frac{1}{2}mv^2}{p_{\text{ph}}c}\)

Since \(p_{\text{ph}} = mv\), replace it to get:

\(\frac{E_e}{E_{\text{ph}}} = \frac{\frac{1}{2}mv^2}{mvc}\)

Canceling same terms, we obtain:

\(\frac{E_e}{E_{\text{ph}}} = \frac{v}{2c}\)

6. Therefore, we conclude that the correct answer is:

\(\frac{E_e}{E_{\text{ph}}} = \frac{v}{2c}\)