JEE Physics 2025 Question with Solution

Given: de Broglie wavelength is $\lambda$ and kinetic energy is $K$ for a particle of constant mass.

Find: the correct graph relating the given quantities.

Using the de Broglie relation,

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

and for a non-relativistic particle,

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

Substituting momentum in the wavelength formula,

$$\lambda = \frac{h}{\sqrt{2mK}}$$

So,

$$\lambda \propto \frac{1}{\sqrt{K}}$$

Squaring both sides,

$$\lambda^2 \propto \frac{1}{K}$$

Hence,

$$\frac{1}{K} \propto \lambda^2$$

Therefore the graph of $\frac{1}{K}$ versus $\lambda$ is an increasing parabola passing through the origin. The correct option is A according to the solution, although the displayed working states the curve should be increasing while option A image appears inconsistent with that description.

Given: $\lambda = \frac{h}{p}$ and mass is constant.

Find: which graph matches the dependence on kinetic energy.

From kinetic energy,

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

So,

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

Now,

$$\lambda = \frac{h}{\sqrt{2mK}}$$

Rearranging,

$$\sqrt{K} = \frac{h}{\lambda\sqrt{2m}}$$

and therefore,

$$K \propto \frac{1}{\lambda^2}$$

So,

$$\frac{1}{K} \propto \lambda^2$$

This means $\frac{1}{K}$ must increase with $\lambda$ in a nonlinear manner. The solution explicitly marks A as correct, so the answer is taken from the solution.