If \( \lambda_e \), \( \lambda_p \) and \( \lambda_\alpha \) be the de-Broglie wavelength of electron, proton and \( \alpha \)-particle, respectively, then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \lambda_\alpha \lt \lambda_p \lt \lambda_e \)
The de Broglie wavelength is \( \lambda = h/p \). The question compares the particles under the same condition, which may be the same speed, the same kinetic energy or the same accelerating potential; all three give the same order.
\( \displaystyle \lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}} \)
Same speed or same kinetic energy: \( \lambda \) decreases as the mass increases, and
\( \displaystyle m_{e} \lt m_{p} \lt m_{\alpha} \)
Same accelerating potential: \( \lambda \propto 1/\sqrt{mq} \), and \( mq \) also increases from electron to proton to \( \alpha \)-particle (\( m_\alpha = 4m_p \), \( q_\alpha = 2e \)).
So the heaviest particle has the shortest wavelength:
\( \displaystyle \lambda_{\alpha} \lt \lambda_{p} \lt \lambda_{e} \)