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 solve this problem, we need to compare the kinetic energy of an electron and a photon given that they have the same de-Broglie wavelength.

First, let's understand the de-Broglie wavelength formula:

\(\lambda = \frac{h}{p}\) 

where \( \lambda \) is the de-Broglie wavelength, \( h \) is Planck's constant, and \( p \) is the momentum.

For an electron, the momentum \( p_e \) is given by:

\(p_e = m v\)

where \( m \) is the mass of the electron and \( v \) is its velocity.

For a photon, the momentum \( p_p \) is:

\(p_p = \frac{E}{c}\)

where \( E \) is the energy of the photon and \( c \) is the speed of light. For a photon, energy \( E = h \nu \) and \( \nu \) is the frequency of the photon.

Given that the velocity of the electron is 25% of the speed of light, we have:

\(v = 0.25c\)

The kinetic energy (K.E.) of the electron is:

K.E._e = \frac{1}{2} m v^2\)

Since the de-Broglie wavelength of the electron and the photon is the same, their momenta are equal:

\(\frac{h}{m v} = \frac{h}{p_p}\)

The kinetic energy of the photon (considering relativistic energy) is purely its energy due to its frequency, which relates to its momentum:

K.E._p = p_p c\)

The ratio of kinetic energies is given by:

\(\frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} m v^2}{p_p c}\)

Replacing \(p_p\) with \(m v\) from the equality of momenta:

\(\frac{K.E._e}{K.E._p} = \frac{\frac{1}{2} m (0.25c)^2}{m (0.25c) c}\)

After simplifying:

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

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

\(\frac{1}{8}\).

Answer 2

Step 1: For the Photon - The energy of a photon \( E_p \) is given by:

\[ E_p = \frac{hc}{\lambda_p} \]

- Rearranging for wavelength \(\lambda_p\), we get:

\[ \lambda_p = \frac{hc}{E_p} \]

Step 2: For the Electron - The de-Broglie wavelength of an electron is given by:

\[ \lambda_e = \frac{h}{m_e v_e} \]

- The kinetic energy \( K_e \) of the electron is related to its velocity by:

\[ K_e = \frac{1}{2} m_e v_e^2 \]

- Rearranging, the velocity \( v_e \) can be expressed as:

\[ v_e = \sqrt{\frac{2 K_e}{m_e}} \]

Step 3: Equating Wavelengths - Since the de-Broglie wavelength of the electron is the same as that of the photon, we equate \(\lambda_p\) and \(\lambda_e\):

\[ \frac{hc}{E_p} = \frac{h}{m_e v_e} \]

- Simplifying, we get:

\[ E_p = m_e v_e c \]

Step 4: Express \( v_e \) in Terms of \( c \) - We are given that \( v_e = 0.25c \). - Substitute \( v_e = 0.25c \) into the expression for \( E_p \):

\[ E_p = m_e (0.25c) c = 0.25 m_e c^2 \]

Step 5: Calculate the Ratio of Kinetic Energies - The kinetic energy of the electron is:

\[ K_e = \frac{1}{2} m_e v_e^2 = \frac{1}{2} m_e (0.25c)^2 = \frac{1}{2} m_e \cdot 0.0625c^2 = 0.03125 m_e c^2 \]

- Now, take the ratio \( \frac{K_e}{E_p} \):

\[ \frac{K_e}{E_p} = \frac{0.03125 m_e c^2}{0.25 m_e c^2} = \frac{1}{8} \]

So, the correct answer is: \( \frac{1}{8} \)