Correct Option: (B) $\lambda_e > \lambda_p > \lambda_d$
Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]
According to the de Broglie hypothesis, the wavelength ($\lambda$) associated with a particle of mass $m$ moving with speed $v$ is given by:
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$Where $h$ is Planck's constant. Given that all particles (electron, proton, and deuteron) move with the same speed $v$, the wavelength is inversely proportional to the mass of the particle:
$$\lambda \propto \frac{1}{m}$$Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]
We compare the masses of the three particles:
- Mass of electron: $m_e \approx 9.1 \times 10^{-31} \text{ kg}$
- Mass of proton: $m_p \approx 1.67 \times 10^{-27} \text{ kg}$
- Mass of deuteron: $m_d \approx 3.34 \times 10^{-27} \text{ kg}$
The relationship between their masses is:
$$m_e < m_p < m_d$$Since $\lambda = \frac{h}{mv}$ and $h, v$ are constant, substituting the mass relationship into the inverse proportionality:
$$\frac{1}{m_e} > \frac{1}{m_p} > \frac{1}{m_d}$$Step 3 (Final Concluded Answer): [1/3 Mark]
Therefore, the wavelengths follow the order:
$$\lambda_e > \lambda_p > \lambda_d$$This corresponds to option (B).
💡 EL10 Concept Insight: Think of the de Broglie wavelength like a "wave-like footprint" left by a moving object; the lighter the object is, the more spread out and wave-like its footprint becomes, which is why the tiny electron has a much larger wavelength than the heavy proton or deuteron.