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Structure of Atom question

2006 · Shift 0 · Q45
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Structure of Atom question

2006 · Shift 0 · Q45

JEE MainChemistryStructure of AtomMCQ+4 / −1
Uncertainty in the position of an electron (mass = 9.1 ×\times× 10-31 kg) moving with a velocity 300 ms-1, accurate upto 0.001% will be (h = 6.63 ×\times× 10-34 Js)
  1. A
    1.92 ×\times× 10-2 m
  2. B
    3.84 ×\times× 10-2 m
  3. C
    19.2 ×\times× 10-2 m
  4. D
    5.76 ×\times× 10-2 m
View written solutionFree

Correct answer: A

  1. Use Heisenberg uncertainty principle

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

    Since momentum p=mvp = mvp=mv, we have:

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

    So,

    Δx=h4πmΔv\Delta x = \frac{h}{4\pi m\Delta v}Δx=4πmΔvh​

  2. Find uncertainty in velocity

    Given velocity v=300 m s−1v = 300\,\text{m s}^{-1}v=300m s−1 accurate up to 0.001%0.001\%0.001%.

    0.001%=0.001100=10−50.001\% = \frac{0.001}{100} = 10^{-5}0.001%=1000.001​=10−5

    Therefore,

    Δv=300×10−5=3×10−3 m s−1\Delta v = 300 \times 10^{-5} = 3 \times 10^{-3}\,\text{m s}^{-1}Δv=300×10−5=3×10−3m s−1

  3. Substitute values

    Given:

    h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s m=9.1×10−31 kgm = 9.1 \times 10^{-31}\,\text{kg}m=9.1×10−31kg Δv=3×10−3 m s−1\Delta v = 3 \times 10^{-3}\,\text{m s}^{-1}Δv=3×10−3m s−1

    Δx=6.63×10−344π×9.1×10−31×3×10−3\Delta x = \frac{6.63 \times 10^{-34}}{4\pi \times 9.1 \times 10^{-31} \times 3 \times 10^{-3}}Δx=4π×9.1×10−31×3×10−36.63×10−34​

  4. Calculate denominator

    9.1×3=27.39.1 \times 3 = 27.39.1×3=27.3

    4π×27.3×10−34≈12.57×27.3×10−344\pi \times 27.3 \times 10^{-34} \approx 12.57 \times 27.3 \times 10^{-34}4π×27.3×10−34≈12.57×27.3×10−34

    ≈343.16×10−34=3.4316×10−32\approx 343.16 \times 10^{-34} = 3.4316 \times 10^{-32}≈343.16×10−34=3.4316×10−32

  5. Now calculate Δx\Delta xΔx

    Δx=6.63×10−343.4316×10−32\Delta x = \frac{6.63 \times 10^{-34}}{3.4316 \times 10^{-32}}Δx=3.4316×10−326.63×10−34​

    =6.633.4316×10−2= \frac{6.63}{3.4316} \times 10^{-2}=3.43166.63​×10−2

    ≈1.93×10−2 m\approx 1.93 \times 10^{-2}\,\text{m}≈1.93×10−2m

  6. Match with options

    Δx≈1.92×10−2 m\Delta x \approx 1.92 \times 10^{-2}\,\text{m}Δx≈1.92×10−2m

    So the correct option is A.

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