An electron, a doubly ionized helium ion $$(He^{++})$$ and proton are having the same kinetic energy. The relation between their respective de-Broglie wavelength $$\lambda_e$$, $$\lambda_{He^{++}}$$ and $$\lambda_p$$ is:

We begin with the de-Broglie relation, which states that for any particle the wavelength associated with its motion is

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

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

Momentum and kinetic energy are related by the classical (non-relativistic) formula

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

where $$K$$ is the kinetic energy and $$m$$ is the mass of the particle. Solving this expression for $$p$$ we have

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

Substituting this expression for $$p$$ back into the de-Broglie formula gives

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

In the given problem the electron $$(e^{-})$$, the proton $$(p)$$ and the doubly ionised helium ion $$(He^{++})$$ are all said to possess the same kinetic energy $$K$$. Because $$h$$ and $$K$$ are common to all three particles, the above formula shows that their wavelengths obey the proportionality

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

Thus, smaller mass $$\Longrightarrow$$ larger de-Broglie wavelength, while larger mass $$\Longrightarrow$$ smaller wavelength.

Now we list their approximate masses:

$$m_e \approx 9.11 \times 10^{-31}\;{\rm kg},$$

$$m_p \approx 1.67 \times 10^{-27}\;{\rm kg},$$

$$m_{He^{++}} \approx 4m_p \approx 6.68 \times 10^{-27}\;{\rm kg}.$$

Clearly

$$m_e \;<\; m_p \;<\; m_{He^{++}}\;.$$

Using the proportionality $$\lambda \propto 1/\sqrt{m}$$, we invert this mass order to obtain the wavelength order:

$$\lambda_e \;>\; \lambda_p \;>\; \lambda_{He^{++}}\;.$$

This matches the option

$$\lambda_e > \lambda_p > \lambda_{He^{++}}.$$

Hence, the correct answer is Option C.