If electron, deuteron, \( \alpha \)-particle and proton have same kinetic energy, which one has the largest de-Broglie wavelength.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → electron
The de Broglie wavelength is \( \lambda = \frac{h}{p} \), and the momentum in terms of the kinetic energy \( K \) is \( p = \sqrt{2mK} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \propto \frac{1}{\sqrt{m}} \quad (\text{same } K) \)
The wavelength is largest for the particle of smallest mass. The electron is about 1800 times lighter than the proton, and the proton, deuteron and \( \alpha \)-particle have masses in the ratio \( 1 : 2 : 4 \) (nearly), so the electron has the largest wavelength.