MCQ
In atom, an electron is moving with a speed of $600m/ s$ with an accuracy of $0.005\%$. Certainty with which the position of the electron can be located is : $\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~kg} / \mathrm{m}^2 \mathrm{~s}^{-1}\right.,$ mass of electron, em $= 6.6 \times 10^{-31}kg)$
  • A
    $1.52 \times 10^{-4}m$
  • B
    $5.10 \times 10^{-3}m$
  • $1.92 \times 10^{-3}m$
  • D
    $3.84 \times 10^{-3}m$

Answer

Correct option: C.
$1.92 \times 10^{-3}m$
By Heisenberg's uncertainty principle,
$\Delta\text{x}.\text{m}\Delta\text{v}=\frac{\text{h}}{4\pi}$
$\Delta\text{v}=0.005\%$ or $600\text{ms}=\frac{600\times0.005}{100}=0.03$
$\Delta\text{x}\times9.1\times10^{-31}\times0.03=\frac{6.6\times10^{-34}}{4\times3.14}$
Hence, $\Delta\text{x}=\frac{6.6\times10^{-34}}{4\times3.14\times0.03\times9.1\times10^{-31}}$
$=1.92\times10^{-3}\text{m}$

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