MCQ
The mass of electron is $9.11 \times 10^{-31} \ kg,$ Planck constant is $6.626 \times 10^{-34} \ J\ s,$ the uncertainty involved in the measurement of velocity within a distance of $0.1 \ \mathop A\limits^o $ is
  • A
    $5.79 \times 10^5 \ m \ s^{-1}$
  • $5.79 \times 10^6 \ m \ s^{-1}$
  • C
    $5.79 \times 10^7 \ m\  s^{-1}$
  • D
    $5.79 \times 10^8\ m\ s^{-1}$

Answer

Correct option: B.
$5.79 \times 10^6 \ m \ s^{-1}$
b
By Heisenberg's uncertainty principle

$\Delta p \cdot \Delta x \geq \frac{h}{2 \pi}$

or $\Delta \mathrm{v} . \Delta x \geq \frac{\mathrm{h}}{4 \pi \mathrm{m}}$

$\Delta \mathrm{p} \rightarrow$ uncertainty in momentum

$\Delta x \rightarrow$ uncertainty in position

$\Delta \mathrm{v} \rightarrow$ uncertainty in velocity

$\mathrm{m} \rightarrow$ mass of particle Given, $\Delta x=0.1 \mathrm{A}=0.1 \times 10^{-10} \mathrm{m}$

$m=9.11 \times 10^{-31} \mathrm{kg}$

$\mathrm{h}=$ planck constant $=6.626 \times 10^{-34} \mathrm{Js}$

In uncertain position $\Delta v \cdot \Delta x=\frac{h}{4 \pi m}$ $\Delta v \times 0.1 \times 10^{-10}=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.11 \times 10^{-31}}$

$\Delta v=\frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 911 \times 10^{-31} \times 0.1 \times 10^{-1}} \mathrm{ms}^{-1}$

$=5.785 \times 10^{6} \mathrm{ms}^{-1}$

$5.79 \times 10^{6} \mathrm{ms}^{-1}$

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