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

Answer: (B)

$\frac{1}{8}$

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: $\frac{1}{8}$