If the de Broglie wavelengths associated with a proton and an $$\alpha$$-particle are equal, then the ratio of velocities of the proton and the $$\alpha$$-particle will be:

We begin with the de Broglie relation, which states that the wavelength $$\lambda$$ associated with any material particle is given by the formula

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

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle. Writing momentum in terms of mass and velocity, we have

$$p \;=\; m v,$$

so that the wavelength becomes

$$\lambda \;=\;\dfrac{h}{m v}.$$

According to the problem, the de Broglie wavelengths of a proton (subscript $$p$$) and an $$\alpha$$-particle (subscript $$\alpha$$) are equal. Hence

$$\lambda_p \;=\; \lambda_\alpha.$$

Substituting the expression for each wavelength, we get

$$\dfrac{h}{m_p v_p} \;=\; \dfrac{h}{m_\alpha v_\alpha}.$$

Because Planck’s constant $$h$$ is the same on both sides, it cancels out, leaving

$$\dfrac{1}{m_p v_p} \;=\; \dfrac{1}{m_\alpha v_\alpha}.$$

Cross-multiplying, we obtain

$$m_\alpha v_\alpha \;=\; m_p v_p.$$

Now we use the known masses. A proton has mass $$m_p$$, while an $$\alpha$$-particle (which consists of two protons and two neutrons) has approximately four times that mass, so

$$m_\alpha \;=\; 4\,m_p.$$

Substituting this value into the momentum equality, we have

$$4\,m_p\,v_\alpha \;=\; m_p\,v_p.$$

Dividing both sides by $$m_p$$ (which is non-zero), we find

$$4\,v_\alpha \;=\; v_p.$$

Rearranging to express the ratio of velocities,

$$\dfrac{v_p}{v_\alpha} \;=\; 4.$$

This tells us that

$$v_p : v_\alpha \;=\; 4 : 1.$$

Hence, the correct answer is Option C.