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Structure of Atom question

2002 · Shift 0 · Q55
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Structure of Atom question

2002 · Shift 0 · Q55

JEE MainChemistryStructure of AtomMCQ+4 / −1
Uncertainty in position of a minute particle of mass 25 g in space is 10-5 m. What is the uncertainty in its velocity (in ms-1) (h = 6.6 ×\times× 10-34 Js)
  1. A
    2.4 ×\times× 10-34
  2. B
    0.5 ×\times× 10-34
  3. C
    2.1 ×\times× 10-28
  4. D
    0.5 ×\times× 10-23
View written solutionFree

Correct answer: C

  1. Use Heisenberg uncertainty principle

    For position and momentum, Delta x\, \Delta p \ge \frac{h}{4\pi}

    Since momentum p=mvp = mvp=mv, Δp=m Δv\Delta p = m\,\Delta vΔp=mΔv

    So, Δx⋅m Δv≥h4π\Delta x \cdot m\,\Delta v \ge \frac{h}{4\pi}Δx⋅mΔv≥4πh​

    Hence, Δv≥h4πmΔx\Delta v \ge \frac{h}{4\pi m \Delta x}Δv≥4πmΔxh​

  2. Substitute the given values

    Given: m=25 g=25×10−3 kg=0.025 kgm = 25\text{ g} = 25 \times 10^{-3}\text{ kg} = 0.025\text{ kg}m=25 g=25×10−3 kg=0.025 kg Δx=10−5 m\Delta x = 10^{-5}\text{ m}Δx=10−5 m h=6.6×10−34 Jsh = 6.6 \times 10^{-34}\text{ Js}h=6.6×10−34 Js

    Therefore, Δv=6.6×10−344π×0.025×10−5\Delta v = \frac{6.6 \times 10^{-34}}{4\pi \times 0.025 \times 10^{-5}}Δv=4π×0.025×10−56.6×10−34​

  3. Simplify the denominator

    0.025×10−5=2.5×10−70.025 \times 10^{-5} = 2.5 \times 10^{-7}0.025×10−5=2.5×10−7

    So, 4π×2.5×10−7=10π×10−7=π×10−64\pi \times 2.5 \times 10^{-7} = 10\pi \times 10^{-7} = \pi \times 10^{-6}4π×2.5×10−7=10π×10−7=π×10−6

    Thus, Δv=6.6×10−34π×10−6\Delta v = \frac{6.6 \times 10^{-34}}{\pi \times 10^{-6}}Δv=π×10−66.6×10−34​ =6.6π×10−28= \frac{6.6}{\pi} \times 10^{-28}=π6.6​×10−28

  4. Numerical value

    Taking π≈3.14\pi \approx 3.14π≈3.14, Δv≈6.63.14×10−28\Delta v \approx \frac{6.6}{3.14} \times 10^{-28}Δv≈3.146.6​×10−28 Δv≈2.1×10−28 m s−1\Delta v \approx 2.1 \times 10^{-28}\text{ m s}^{-1}Δv≈2.1×10−28 m s−1

  5. Match with options

    This corresponds to: Option C: 2.1×10−282.1 \times 10^{-28}2.1×10−28

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