An electron, an \( \alpha \) particle, a proton and a deuteron 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 \( \lambda = h/p \). A particle of mass \( m \) and kinetic energy \( K \) has momentum \( p = \sqrt{2mK} \), so
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \)
For the same kinetic energy, \( \lambda \propto 1/\sqrt{m} \): the heaviest particle has the shortest wavelength.
Taking the proton mass as \( m_p \), the masses are approximately
\( \text{electron: } m_p/1836 \)
\( \text{proton: } m_p \)
\( \text{deuteron: } 2m_p \)
\( \alpha\text{ particle: } 4m_p \)
The \( \alpha \) particle is the heaviest, so its de Broglie wavelength is the shortest.