A moving proton and electron have the same de-Broglie wavelength. If $$K$$ and $$P$$ denote the K.E. and momentum respectively. Then choose the correct option:

We begin with the de-Broglie relation, which states that the wavelength $$\lambda$$ of any moving particle is connected to its linear momentum $$P$$ through the formula $$\lambda=\dfrac{h}{P}$$, where $$h$$ is Planck’s constant.

The problem tells us that the proton and the electron have the same de-Broglie wavelength. Using the above relation, we write for the proton (subscript $$p$$) and for the electron (subscript $$e$$)

$$\lambda_p=\dfrac{h}{P_p}\quad\text{and}\quad\lambda_e=\dfrac{h}{P_e}$$.

Since the wavelengths are equal, $$\lambda_p=\lambda_e$$, so

$$\dfrac{h}{P_p}=\dfrac{h}{P_e}.$$

Canceling the common factor $$h$$ on both sides, we get

$$P_p=P_e.$$

Hence the momenta of the proton and the electron are identical.

Next, we compare their kinetic energies. For non-relativistic motion, the kinetic energy $$K$$ of a particle of mass $$m$$ and momentum $$P$$ is given by the formula

$$K=\dfrac{P^{2}}{2m}.$$

Applying this to each particle, we have

$$K_p=\dfrac{P_p^{2}}{2m_p}\quad\text{and}\quad K_e=\dfrac{P_e^{2}}{2m_e}.$$

We have just shown $$P_p=P_e$$. Let us denote this common value simply by $$P$$. Substituting,

$$K_p=\dfrac{P^{2}}{2m_p},\qquad K_e=\dfrac{P^{2}}{2m_e}.$$

Now, the rest mass of a proton $$m_p$$ is much larger than the rest mass of an electron $$m_e$$; numerically, $$m_p\approx1836\,m_e$$. Because the denominator in $$K_p$$ is larger, the value of $$K_p$$ is smaller. Thus,

$$K_p<K_e.$$

Combining our two conclusions,

$$K_p<K_e\quad\text{and}\quad P_p=P_e.$$

This matches Option B in the given list.

Hence, the correct answer is Option B.