If the uncertainty in velocity and position of a minute particle in space are, 2.4 \times 10 - 26 (m s - 1) and 10 - 7 (m) respectively. The mass of the particle in g is \underline\hspace2cm . (Nearest integer) (Given : h = 6.626 \times 10 - 34 Js)

Structure of Atom PYQ — JEE Main 2022 · 27 Jun · Shift 1 · Q11 with Solution

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Correct answer: 22

For position and momentum,

\Delta x\,\Delta p \ge \frac{h}{4\pi}

Since

\Delta p = m\,\Delta v

we get

\Delta x\, m\, \Delta v \ge \frac{h}{4\pi}

So,

m \ge \frac{h}{4\pi\,\Delta x\,\Delta v}

Given:

\Delta v = 2.4 \times 10^{-26}\ \text{m s}^{-1}

\Delta x = 10^{-7}\ \text{m}

h = 6.626 \times 10^{-34}\ \text{J s}

Thus,

m = \frac{6.626 \times 10^{-34}}{4\pi \times 10^{-7} \times 2.4 \times 10^{-26}}

10^{-7} \times 10^{-26} = 10^{-33}

So,

m = \frac{6.626 \times 10^{-34}}{4\pi \times 2.4 \times 10^{-33}}

m = \frac{6.626}{9.6\pi} \times 10^{-1}

Using $\pi \approx 3.1416$ ,

9.6\pi \approx 30.159

Therefore,

m \approx \frac{6.626}{30.159} \times 10^{-1}

m \approx 0.2197 \times 10^{-1}

m \approx 2.197 \times 10^{-2}\ \text{kg}

2.197 \times 10^{-2}\ \text{kg} = 21.97\ \text{g}

Nearest integer:

\boxed{22}

Stored correct answer = $22$

Our derived answer = $22$

So, the answer agrees with the stored correct answer.