JEE Physics 2025 Question with Solution

Given: A proton of mass $m_p$ has the same energy as a photon of wavelength $\lambda$. The proton moves at non-relativistic speed.

Find: The ratio $\frac{\lambda_p}{\lambda}$.

For the photon,

$$E = \frac{hc}{\lambda}$$

so

$$\lambda = \frac{hc}{E}$$

For the proton, the de Broglie wavelength is

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

Its kinetic energy is

$$E = \frac{p^2}{2m_p}$$

Therefore,

$$p^2 = 2m_pE$$

and

$$p = \sqrt{2m_pE}$$

Substituting into the de Broglie relation,

$$\lambda_p = \frac{h}{\sqrt{2m_pE}}$$

Now,

$$\frac{\lambda_p}{\lambda} = \frac{\frac{h}{\sqrt{2m_pE}}}{\frac{hc}{E}}$$ $$= \frac{h}{\sqrt{2m_pE}} \times \frac{E}{hc}$$ $$= \frac{E}{c\sqrt{2m_pE}}$$ $$= \frac{\sqrt{E}}{c\sqrt{2m_p}}$$ $$= \frac{1}{c}\sqrt{\frac{E}{2m_p}}$$

Therefore, the required ratio is $\frac{1}{c}\sqrt{\frac{E}{2m_p}}$. The solution marks the correct option as A.

Given: The energy of the proton equals the energy of a photon of wavelength $\lambda$.

Find: $\frac{\lambda_p}{\lambda}$.

Photon energy is

$$E = \frac{hc}{\lambda}$$

For a non-relativistic proton,

$$E = \frac{1}{2}m_pv^2$$

Since both energies are equal,

$$\frac{1}{2}m_pv^2 = \frac{hc}{\lambda}$$

Thus,

$$v = \sqrt{\frac{2hc}{m_p\lambda}}$$

Now the de Broglie wavelength of the proton is

$$\lambda_p = \frac{h}{m_pv}$$

Substituting $v$,

$$\lambda_p = \frac{h}{m_p\sqrt{\frac{2hc}{m_p\lambda}}}$$ $$= \sqrt{\frac{h\lambda}{2m_pc}}$$

Hence,

$$\frac{\lambda_p}{\lambda} = \frac{\sqrt{\frac{h\lambda}{2m_pc}}}{\lambda}$$ $$= \sqrt{\frac{h}{2m_pc\lambda}}$$

Using

$$E = \frac{hc}{\lambda}$$

we get

$$\frac{h}{\lambda} = \frac{E}{c}$$

Therefore,

$$\frac{\lambda_p}{\lambda} = \frac{1}{c}\sqrt{\frac{E}{2m_p}}$$

Therefore, the correct option is A. Note that the expression written in the options is not algebraically identical to the simplified final form shown in the solution, but the solution explicitly identifies A as correct.