Question

Let \(E_{e}\) and \(E_{p}\) represent kinetic energy of electron and photon respectively. If de-Broglie wavelength \(\lambda _{p}\) of a photon is four times the de-Broglie wavelength \(\lambda _{e}\) of an electron, then \(\frac{E_{p}}{E_{e}}\) is
[ speed of electron \(=\frac{C}{100}\) , C = velocity of light]

• A. \(\frac{1}{100}\)
• B. \(\frac{1}{50}\)
• C. \(50\)
• D. \(100\)

Hint:

Write the photon energy as hc over wavelength and the electron kinetic energy through its de Broglie wavelength, then use the given ratio of wavelengths.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Photon energy is \(E_p=\dfrac{hc}{\lambda_p}\).
For the electron, \(\lambda_e=\dfrac{h}{mv}\) and \(E_e=\dfrac{1}{2}mv^{2}\).

Step 2: Electron Energy
From \(\lambda_e=\dfrac{h}{mv}\) we get \(mv=\dfrac{h}{\lambda_e}\), so \(E_e=\dfrac{1}{2}v\,(mv)=\dfrac{hv}{2\lambda_e}\).

Step 3: Ratio
\(\dfrac{E_p}{E_e}=\dfrac{hc/\lambda_p}{hv/(2\lambda_e)}=\dfrac{2c}{v}\cdot\dfrac{\lambda_e}{\lambda_p}\).

Step 4: Calculation
Put \(v=\dfrac{c}{100}\) and \(\lambda_p=4\lambda_e\):
\(\dfrac{E_p}{E_e}=2\times 100\times\dfrac{1}{4}=50\).

Final Answer:
The ratio of photon energy to electron kinetic energy is 50. \[ \boxed{\text{(C) }50} \]