JEE Physics 2023 Question with Solution

Given: The proton and the alpha particle have the same de-Broglie wavelength $\lambda$.

Find: The ratio $\dfrac{KE_p}{KE_\alpha}$.

For a particle,

$$KE = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$$

Since both particles have the same $\lambda$, and $h$ is constant, kinetic energy is inversely proportional to mass:

$$KE \propto \frac{1}{m}$$

Therefore,

$$\frac{KE_p}{KE_\alpha} = \frac{m_\alpha}{m_p}$$

For an alpha particle,

$$m_\alpha = 4m_p$$

So,

$$\frac{KE_p}{KE_\alpha} = 4:1$$

Therefore, the correct option is B.

Given: Same de-Broglie wavelength $\lambda$ for both particles.

Find: Ratio of kinetic energies.

De-Broglie wavelength is given by

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

If $\lambda$ is the same for both particles, then their momenta are equal.

Now use

$$KE = \frac{p^2}{2m}$$

With the same momentum $p$, the particle with smaller mass has larger kinetic energy. Hence,

$$\frac{KE_p}{KE_\alpha} = \frac{p^2/2m_p}{p^2/2m_\alpha} = \frac{m_\alpha}{m_p}$$

Since the mass of the alpha particle is four times the mass of the proton,

$$\frac{KE_p}{KE_\alpha} = 4:1$$

Hence, the correct option is B.