JEE Main 2025 — Structure Of Atom Question with Solution

Statement I : Correct

This statement is describing the uncertainty principle.

It states that it is impossible to measure both the position and the momentum of an object.

Statement II : Correct.

If the uncertainty in the measurement of position ($\Delta x$) and the uncertainty in the measurement of momentum ($\Delta p$) are equal for an electron, then the uncertainty in the measurement of velocity ($\Delta v$) is greater than or equal to $\sqrt{\frac{h}{\pi }}\frac{1}{2m}$

This is derived from the Heisenberg uncertainty principle, $$\Delta x.\Delta p_x\ge \frac{h}{4\pi }$$

Given, $$\Delta x=\Delta p$$

Substitute this in $$\Delta x.\Delta p\ge \frac{h}{4\pi }$$ as

$$\Delta p.\Delta p\ge \frac{h}{4\pi }$$

$$\Delta p^2\ge \frac{h}{4\pi }$$

$$\Delta p^2\ge \frac{h}{4\pi }$$

$$\{\text{matrix}Theformulaformomentum(P)isP=mV\text{hfill}\text{cr}Foruncerta\mathrm{int}y\Delta P=m\Delta v\text{hfill}\text{cr}$$

So, $$m\Delta v\ge \sqrt{\frac{h}{4\pi }}$$

$$\Delta v\ge \sqrt{\frac{h}{4\pi }}\times \frac{1}{m}$$

$$\Delta v\ge \frac{1}{2}>\sqrt{\frac{h}{\pi }}>\frac{1}{m}$$

$$\Delta v\ge \sqrt{\frac{h}{\pi }}\frac{1}{2m}$$

Both the statements are true (correct). So answer is (2) Both statement I and statement II are true.