Question

The de Broglie wavelength of a charged particle accelerated through a potential difference $ V $ is $ \lambda $. If the potential difference is increased by 21%, the de Broglie wavelength of the charged particle is:

• A. \( \frac{5\lambda}{9} \)
• B. \( \frac{7\lambda}{9} \)
• C. \( \frac{9\lambda}{11} \)
• D. \( \frac{10\lambda}{11} \)

Hint:

For de Broglie waves, the wavelength is inversely proportional to the square root of the potential difference. Any increase in potential difference will decrease the de Broglie wavelength.

Answer 1

The Correct Option is D

1. We know that the de Broglie wavelength \( \lambda \) is related to the potential difference \( V \) by the equation: \[ \lambda = \frac{h}{\sqrt{2meV}} \] Where:
\( h \) is Planck's constant,
\( m \) is the mass of the particle,
\( e \) is the charge of the particle,
\( V \) is the potential difference.
2. If the potential difference \( V \) is increased by 21%, the new potential difference becomes \( 1.21V \).
3. Now, the de Broglie wavelength for the new potential difference is: \[ \lambda' = \frac{h}{\sqrt{2me(1.21V)}} \] 4. The ratio of the new wavelength \( \lambda' \) to the original wavelength \( \lambda \) is: \[ \frac{\lambda'}{\lambda} = \frac{\sqrt{2meV}}{\sqrt{2me(1.21V)}} = \frac{1}{\sqrt{1.21}} \approx \frac{1}{1.1} \] 5. Thus, the new wavelength is: \[ \lambda' = \frac{\lambda}{1.1} \approx \frac{10\lambda}{11} \] 6. Therefore, the correct answer is \( \frac{10\lambda}{11} \), which corresponds to option (4).