What should be the order of arrangement of de-Broglie wavelength of electron $$\lambda_e$$, an $$\alpha$$-particle $$\lambda_\alpha$$ and proton $$\lambda_p$$ given that all have the same kinetic energy?

We begin with the de-Broglie relation, which connects the wavelength $$\lambda$$ of a particle with its linear momentum $$p$$:

$$\lambda \;=\; \frac{h}{p},$$

where $$h$$ is Planck’s constant.

All the three particles mentioned—electron, proton and $$\alpha$$-particle—are said to possess the same kinetic energy. For a non-relativistic particle, the kinetic energy $$K$$ and the momentum $$p$$ are related through the familiar formula from elementary mechanics:

$$K \;=\; \frac{p^{2}}{2m},$$

where $$m$$ denotes the mass of the particle. We now solve this expression for the momentum $$p$$ because the de-Broglie wavelength depends directly on $$p$$.

Multiplying both sides by $$2m$$ gives

$$2mK \;=\; p^{2}.$$

Taking the positive square root (momentum magnitude) on both sides, we get

$$p \;=\; \sqrt{\,2mK\,}.$$

We now substitute this explicit form of $$p$$ into the de-Broglie relation:

$$\lambda \;=\; \frac{h}{p} \;=\; \frac{h}{\sqrt{\,2mK\,}}.$$

Because all three particles are given to have the same kinetic energy $$K$$, the symbols $$h$$ and $$K$$ are identical for each. Therefore, the only variable determining the relative size of the wavelength is the mass $$m$$ present under the square root in the denominator.

Let us state this dependence clearly:

$$\lambda \,\propto\, \frac{1}{\sqrt{m}}.$$

That is, the de-Broglie wavelength is inversely proportional to the square root of the particle’s mass. A smaller mass yields a larger wavelength, and a larger mass yields a smaller wavelength.

Now we list the approximate masses of the three particles involved:

• Electron mass: $$m_e \approx 9.1 \times 10^{-31}\ \text{kg}.$$
• Proton mass: $$m_p \approx 1.67 \times 10^{-27}\ \text{kg}.$$
• $$\alpha$$-particle mass: $$m_\alpha \approx 4\,m_p \approx 6.68 \times 10^{-27}\ \text{kg}.$$

Plainly,

$$m_e \;<\; m_p \;<\; m_\alpha.$$

Because the wavelength varies as the inverse square root of the mass, the inequality in wavelengths runs in the opposite sense:

$$\lambda_e \;>\; \lambda_p \;>\; \lambda_\alpha.$$

Translating this into the order requested, we have

$$\lambda_e \;>\; \lambda_p \;>\; \lambda_\alpha.$$

Among the given choices, this ordering corresponds precisely to Option C.

Hence, the correct answer is Option C.