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Chemistry · Structure of Atom

JEE Main 2024 — 9 April, Shift 2 — Question 79

Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10−15 m10^{-15}\ \mathrm{m} is x×109 m s−1x \times 10^{9}\ \mathrm{m\,s^{-1}} (nearest integer).

[Given: mass of electron =9.1×10−31 kg= 9.1 \times 10^{-31}\ \mathrm{kg}, Planck's constant h=6.626×10−34 J sh = 6.626 \times 10^{-34}\ \mathrm{J\,s}, π=3.14\pi = 3.14]

Answer: 58

Numerical answer — enter this value.

Step-by-step solution

Heisenberg uncertainty principle, ΔxΔp≥h4π\mathrm{\Delta x \Delta p \ge \frac{h}{4\pi}}

Δp=mΔv\mathrm{\Delta p = m \Delta v} Δv≥h4πmΔx\mathrm{\Delta v \ge \frac{h}{4\pi m \Delta x}}

On substituting values, we have

Δv=6.626×10−344×3.14×9.1×10−31×10−15\mathrm{\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}}} Δv≈5.8×1010 m s−1\mathrm{\Delta v \approx 5.8 \times 10^{10}\ m\,s^{-1}} Δv=58×109 m s−1\mathrm{\Delta v = 58 \times 10^{9}\ m\,s^{-1}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Chemistry
Chapter
Structure of Atom
Topic
Wave-Particle Duality of Matter - de Broglie, Heisenberg

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