An electron, an \( \alpha \)-particle, a proton and a neutron have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \alpha \)-particle
The de Broglie wavelength is \( \lambda = h/p \). For a particle of mass \( m \) and kinetic energy \( K \), the momentum is \( p = \sqrt{2mK} \), so
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)
For the same kinetic energy, \( \lambda \propto 1/\sqrt{m} \): the heaviest particle has the shortest wavelength.
\( \displaystyle m_{e} \lt m_{p} \approx m_{n} \lt m_{\alpha} \approx 4m_{p} \)
So the \( \alpha \)-particle has the shortest de Broglie wavelength, and the electron the longest.