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Structure of Atom question

2021 · 26 Feb · Shift 2 · Q24
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Structure of Atom question

2021 · 26 Feb · Shift 2 · Q24

JEE MainChemistryStructure of AtomNumerical+4 / −1
A ball weighing 10 g is moving with a velocity of 90 ms −-− 1. If the uncertainty in its velocity is 5%, then the uncertainty in its position is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 33 m. (Rounded off to the nearest integer) [Given : h = 6.63 ×\times× 10 −-− 34 Js]
Numerical answer
View written solutionFree

Correct answer: 1

  1. Given data
  • Mass of ball: m=10 g=0.01 kgm = 10\,\text{g} = 0.01\,\text{kg}m=10g=0.01kg
  • Velocity: v=90 m s−1v = 90\,\text{m s}^{-1}v=90m s−1
  • Uncertainty in velocity = 5%5\%5%
  • Planck's constant: h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s

We need uncertainty in position using Heisenberg's uncertainty principle.

  1. Find uncertainty in velocity

Δv=5% of 90=5100×90=4.5 m s−1\Delta v = 5\% \text{ of } 90 = \frac{5}{100} \times 90 = 4.5\,\text{m s}^{-1}Δv=5% of 90=1005​×90=4.5m s−1

  1. Find uncertainty in momentum

Since Δp=m Δv\Delta p = m\,\Delta vΔp=mΔv

So, Δp=0.01×4.5=0.045 kg m s−1\Delta p = 0.01 \times 4.5 = 0.045\,\text{kg m s}^{-1}Δp=0.01×4.5=0.045kg m s−1

  1. Apply Heisenberg uncertainty principle

Δx Δp≥h4π\Delta x \, \Delta p \ge \frac{h}{4\pi}ΔxΔp≥4πh​

Thus, Δx=h4π Δp\Delta x = \frac{h}{4\pi\,\Delta p}Δx=4πΔph​

Substitute values: Δx=6.63×10−344π×0.045\Delta x = \frac{6.63 \times 10^{-34}}{4\pi \times 0.045}Δx=4π×0.0456.63×10−34​

Now, 4π×0.045≈12.566×0.045≈0.5654\pi \times 0.045 \approx 12.566 \times 0.045 \approx 0.5654π×0.045≈12.566×0.045≈0.565

Therefore, Δx≈6.63×10−340.565≈1.17×10−33 m\Delta x \approx \frac{6.63 \times 10^{-34}}{0.565} \approx 1.17 \times 10^{-33}\,\text{m}Δx≈0.5656.63×10−34​≈1.17×10−33m

  1. Round to nearest integer in the form

Δx=(1.17)×10−33 m\Delta x = (1.17) \times 10^{-33}\,\text{m}Δx=(1.17)×10−33m

Rounded to nearest integer: 111

So the required answer is: 1\boxed{1}1​

  1. Comparison with stored correct answer

Stored correct answer = 111

This matches our derived answer.

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