JEE Physics 2023 Question with Solution

Given: Proton and electron have the same kinetic energy.

Find: The ratio of their de-Broglie wavelengths.

For a particle,

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

and momentum can be related to kinetic energy using

$$K = \frac{1}{2}mv^2$$

So,

$$v = \sqrt{\frac{2K}{m}}$$

Hence,

$$p = mv = m\sqrt{\frac{2K}{m}} = \sqrt{2mK}$$

Therefore de-Broglie wavelength is inversely proportional to $\sqrt{m}$ when kinetic energy is same.

So,

$$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p}}$$

Using the solution result $m_p = 1849\,m_e$,

$$\frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{1849m_e}} = \frac{1}{43}$$

Thus the ratio of de-Broglie wavelengths is $1:43$.

The correct option is C.

Given: Same kinetic energy for proton and electron.

Find: $\lambda_p : \lambda_e$

From de-Broglie relation,

$$\frac{\lambda_p}{\lambda_e} = \frac{p_e}{p_p}$$

Now,

$$p = mv$$

and from kinetic energy,

$$K = \frac{1}{2}mv^2$$

So for proton and electron respectively,

$$v_p = \sqrt{\frac{2K}{m_p}}, \qquad v_e = \sqrt{\frac{2K}{m_e}}$$

Substituting into the momentum ratio,

$$\frac{\lambda_p}{\lambda_e} = \frac{m_e v_e}{m_p v_p}$$ $$= \frac{m_e \sqrt{\frac{2K}{m_e}}}{m_p \sqrt{\frac{2K}{m_p}}}$$ $$= \sqrt{\frac{m_e}{m_p}}$$

Using the worked solution value $m_p = 1849m_e$,

$$\frac{\lambda_p}{\lambda_e} = \frac{1}{43}$$

So the correct ratio is $1:43$.

Note: The question statement mentions $1840$, but the provided solution uses $1849$ to obtain the matching option.