A proton, electron and an \( \alpha \) particle are accelerated with same potential. Arrange their de-Broglie wavelengths in ascending order.
A. Wavelength of proton = \( (\lambda_p) \)
B. Wavelength of electron = \( (\lambda_e) \)
C. Wavelength of particle = \( (\lambda_\alpha) \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \lambda_\alpha \lt \lambda_p \lt \lambda_e \)
A charge \( q \) accelerated through a potential \( V \) gains kinetic energy \( K = qV \), so its momentum is \( p = \sqrt{2mqV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}} \)
For the same \( V \), \( \lambda \propto 1/\sqrt{mq} \): the larger the product \( mq \), the shorter the wavelength.
\( \text{electron: } mq = m_e e \)
\( \text{proton: } mq = m_p e \)
\( \alpha\text{ particle: } mq = (4m_p)(2e) = 8m_p e \)
Since \( m_e \ll m_p \), the products are in the order \( m_e e \lt m_p e \lt 8m_p e \). The wavelengths are in the reverse order:
\( \lambda_\alpha \lt \lambda_p \lt \lambda_e \)