Question

The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:

• A. \( \frac{1}{1} \)
• B. \( \frac{1}{8} \)
• C. 8 : 1
• D. \( \frac{1}{4} \)

Answer 1

The Correct Option is B

To determine the kinetic energy ratio between an electron and a photon with identical de-Broglie wavelengths, we first examine the de-Broglie wavelength formula: \( \lambda = \frac{h}{p} \). Here, \( \lambda \) represents the de-Broglie wavelength, \( h \) is Planck's constant, and \( p \) is momentum.

For an electron, momentum \( p_e \) is expressed as \( p_e = m v \), where \( m \) is the electron's mass and \( v \) is its velocity. For a photon, momentum \( p_p \) is given by \( p_p = \frac{E}{c} \), with \( E \) being the photon's energy and \( c \) the speed of light. The photon's energy is also related to its frequency \( u \) by \( E = h u \).

Given that the electron's velocity is 25% of the speed of light, \( v = 0.25c \).

The kinetic energy of the electron is \( K.E._e = \frac{1}{2} m v^2 \).

Since the de-Broglie wavelengths are equal, their momenta are also equal: \( \frac{h}{m v} = \frac{h}{p_p} \).

The kinetic energy of a photon, considering its relativistic nature, is equivalent to its energy related to momentum: \( K.E._p = p_p c \).

The ratio of kinetic energies is \( \frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} m v^2}{p_p c} \).

Substituting \( p_p = m v \) (due to equal momenta) into the ratio yields: \( \frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} m (0.25c)^2}{m (0.25c) c} \).

Simplifying this expression gives: \( \frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} \times 0.0625m c^2}{0.25m c^2} = \frac{1}{8} \).

Therefore, the ratio of the kinetic energy of the electron to that of the photon is \( \frac{1}{8} \). The correct answer is \( \frac{1}{8} \).