Question

An electron moving with speed \( v \) and a photon with speed \( c \), have the same de-Broglie wavelength. The ratio of kinetic energy of the electron to that of the photon is:

• A. \( \frac{3c}{v} \)
• B. \( \frac{v}{3c} \)
• C. \( \frac{v}{2c} \)
• D. \( \frac{2c}{v} \)

Hint:

For particles with the same de-Broglie wavelength, relate the kinetic energy of the particle to the energy of the photon using the de-Broglie wavelength and energy formulas.

Answer 1

The Correct Option is C

Step 1: The de-Broglie wavelength \( \lambda \) of a particle is given by: \[ \lambda = \frac{h}{mv}, \] where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity. 
Step 2: For a photon, the de-Broglie wavelength is related to its momentum \( p \) by: \[ \lambda = \frac{h}{p} = \frac{h}{E/c}, \] where \( E \) is the energy of the photon and \( c \) is the speed of light. 
Step 3: Given that the electron and photon have the same de-Broglie wavelength, we equate the expressions for \( \lambda \) for both: \[ \frac{h}{mv} = \frac{h}{E/c}. \] Simplifying: \[ mv = E/c. \] 
Step 4: The energy of the electron is its kinetic energy \( K_e = \frac{1}{2}mv^2 \), and the energy of the photon is \( E = hf \), where \( f \) is the frequency of the photon. 
Step 5: From the relation \( mv = E/c \), the ratio of the kinetic energy of the electron to the energy of the photon can be written as: \[ \frac{K_e}{E} = \frac{v}{2c}. \]