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Structure of Atom question

2022 · 29 Jul · Shift 1 · Q17
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Structure of Atom question

2022 · 29 Jul · Shift 1 · Q17

JEE MainChemistryStructure of AtomNumerical+4 / −1
The minimum uncertainty in the speed of an electron in an one dimensional region of length 2ao2 \mathrm{a}_{\mathrm{o}}2ao​(Where ao=\mathrm{a}_{\mathrm{o}}=ao​= Bohr radius 52.9 pm52.9 \,\mathrm{pm}52.9pm) is ‾km s−1\underline{\hspace{2cm}}\mathrm{km} \,\mathrm{s}^{-1}​kms−1. (Given : Mass of electron = 9.1 ×\times× 10 −-− 31 kg, Planck's constant h = 6.63 ×\times× 10 −-− 34 Js)
Numerical answer
View written solutionFree

Correct answer: 548

  1. Use Heisenberg uncertainty principle

For position and momentum,

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

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

Δx m Δv≥h4π\Delta x\,m\,\Delta v \ge \frac{h}{4\pi}ΔxmΔv≥4πh​

So the minimum uncertainty in speed is

Δvmin⁡=h4πmΔx\Delta v_{\min} = \frac{h}{4\pi m\Delta x}Δvmin​=4πmΔxh​
  1. Given data

Bohr radius,

a0=52.9 pm=52.9×10−12 ma_0 = 52.9\,\text{pm} = 52.9\times 10^{-12}\,\text{m}a0​=52.9pm=52.9×10−12m

Region length,

2a0=2×52.9×10−12=105.8×10−12=1.058×10−10 m2a_0 = 2\times 52.9\times 10^{-12} = 105.8\times 10^{-12} = 1.058\times 10^{-10}\,\text{m}2a0​=2×52.9×10−12=105.8×10−12=1.058×10−10m

Thus,

Δx=2a0=1.058×10−10 m\Delta x = 2a_0 = 1.058\times 10^{-10}\,\text{m}Δx=2a0​=1.058×10−10m

Also,

m=9.1×10−31 kg,h=6.63×10−34 J sm = 9.1\times 10^{-31}\,\text{kg}, \qquad h = 6.63\times 10^{-34}\,\text{J s}m=9.1×10−31kg,h=6.63×10−34J s
  1. Substitute into the formula
Δvmin⁡=6.63×10−344π×9.1×10−31×1.058×10−10\Delta v_{\min} = \frac{6.63\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 1.058\times 10^{-10}}Δvmin​=4π×9.1×10−31×1.058×10−106.63×10−34​

First calculate the denominator:

9.1×1.058=9.62789.1\times 1.058 = 9.62789.1×1.058=9.6278 4π≈12.5664\pi \approx 12.5664π≈12.566 12.566×9.6278≈121.0112.566\times 9.6278 \approx 121.0112.566×9.6278≈121.01

So denominator is

121.01×10−41=1.2101×10−39121.01\times 10^{-41} = 1.2101\times 10^{-39}121.01×10−41=1.2101×10−39

Hence,

Δvmin⁡≈6.63×10−341.2101×10−39\Delta v_{\min} \approx \frac{6.63\times 10^{-34}}{1.2101\times 10^{-39}}Δvmin​≈1.2101×10−396.63×10−34​ =5.48×105 m s−1= 5.48\times 10^5\,\text{m s}^{-1}=5.48×105m s−1
  1. Convert to km s−1^{-1}−1
5.48×105 m s−1=548 km s−15.48\times 10^5\,\text{m s}^{-1} = 548\,\text{km s}^{-1}5.48×105m s−1=548km s−1
  1. Final answer
548\boxed{548}548​
  1. Comparison with stored answer

Stored correct answer = 548548548

This matches the derived answer exactly.

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