A proton, a deutron, an electron and α-particle have same energy. Their deBroglie wavelengths are \( \lambda_p,\ \lambda_d,\ \lambda_e \) and \( \lambda_\alpha \) respectively.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \lambda_\alpha \lt \lambda_d \lt \lambda_p \lt \lambda_e \)
The de Broglie wavelength of a particle of mass m and kinetic energy K is \( \lambda = h/p \) with \( p = \sqrt{2mK} \).
\( \displaystyle \lambda = \frac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \frac{1}{\sqrt{m}} \quad (\text{same } K) \)
The charge does not enter, so only the masses matter. With \( m_p \) the mass of a proton, the deuteron is about \( 2m_p \) and the α particle about \( 4m_p \), while the electron is far lighter than the proton:
\( m_e \lt m_p \lt m_d \lt m_\alpha \)
The heavier the particle, the shorter its wavelength:
\( \lambda_\alpha \lt \lambda_d \lt \lambda_p \lt \lambda_e \)
Numerically, \( \lambda_p : \lambda_d : \lambda_\alpha = 1 : \frac{1}{\sqrt{2}} : \frac{1}{2} \).