Question

An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength of the electron? (Assume $ h = 6.626 \times 10^{-34}$  J·s , $ m_e = 9.1 \times 10^{-31}$  $\text{kg} $, $ e = 1.6 \times 10^{-19}$ $\,$ $\text{C} $.)

• A. \( 0.061 \, \text{nm} \)
• B. \( 0.123 \, \text{nm} \)
• C. \( 0.246 \, \text{nm} \)
• D. \( 0.492 \, \text{nm} \)

Hint:

The de Broglie wavelength of an electron accelerated through a potential \( V \) can be approximated as \( \lambda \approx \frac{1.226}{\sqrt{V}} \, \text{nm} \).

Answer 1

The Correct Option is B

Step 1: Calculate the kinetic energy. \[ K = eV = (1.6 \times 10^{-19}) \times 100 = 1.6 \times 10^{-17} \, \text{J} \] 
Step 2: Find the momentum. $p = \sqrt{2mK}$ $= \sqrt{2 (9.1 \times 10^{-31}} (1.6 \times 10^{-17})$ $\approx 5.4 \times 10^{-24} \, \text{kgm/s}$ 
Step 3: Calculate the de Broglie wavelength. \[ \lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{5.4 \times 10^{-24}} \approx 1.23 \times 10^{-10} \, \text{m} = 0.123 \, \text{nm} \] 
Step 4: Alternative method. 
For electrons: \[ \lambda (\text{nm}) = \frac{1.226}{\sqrt{V}} = \frac{1.226}{\sqrt{100}} = 0.1226 \, \text{nm} \approx 0.123 \, \text{nm} \] 
Final Answer: \[ \boxed{0.123} \]