Question

The de-Broglie wavelength of an electron moving with a velocity \( \frac{c}{3} \) (where \( c = 3 \times 10^8 \, {m/s} \)) is equal to the wavelength of a photon. The ratio of the kinetic energies of the electron and the photon is:

• A. \( 1 : 4 \)
• B. \( 1 : 3 \)
• C. \( 1 : 2 \)
• D. \( 2 : 1 \)

Hint:

For de-Broglie wavelength and energy calculations, remember the key relationships: - \( \lambda = \frac{h}{mv} \) for particles, - \( E = \frac{hc}{\lambda} \) for photons. The kinetic energy comparison is based on these relationships.

Answer 1

The Correct Option is B

Step 1: The de-Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{mv} \] where \( h \) is Planck's constant and \( m \) is the mass of the particle, and \( v \) is its velocity. For the electron, we have: \[ \lambda_{{electron}} = \frac{h}{m_{{e}} \cdot \frac{c}{3}} \] where \( m_{{e}} \) is the mass of the electron. 
Step 2: For the photon, the wavelength is related to its energy by: \[ E_{{photon}} = \frac{hc}{\lambda_{{photon}}} \] Since the wavelengths of the electron and the photon are equal, we can equate their expressions: \[ \frac{h}{m_{{e}} \cdot \frac{c}{3}} = \frac{hc}{E_{{photon}}} \] This leads to the energy of the photon: \[ E_{{photon}} = \frac{3m_{{e}} c^2}{2} \] 
Step 3: The kinetic energy of the electron is given by: \[ E_{{electron}} = \frac{1}{2} m_{{e}} v^2 = \frac{1}{2} m_{{e}} \left( \frac{c}{3} \right)^2 = \frac{1}{2} m_{{e}} \cdot \frac{c^2}{9} \] 
Step 4: The ratio of the kinetic energies of the electron and the photon is: \[ {Ratio} = \frac{E_{{electron}}}{E_{{photon}}} = \frac{\frac{1}{2} m_{{e}} \cdot \frac{c^2}{9}}{\frac{3m_{{e}} c^2}{2}} = \frac{1}{3} \] 
Step 5: Hence, the ratio of the kinetic energies is \( 1 : 3 \).