Calculate the mass of an elementary particle, which is accelerated to twice the velocity of light with the…

Calculate the mass of an elementary particle, which is accelerated to twice the velocity of light with the precision $\pm 1 \%$ and has $1.05 \times 10^{-13} \mathrm{~m}$ uncertainty in position. $\left(h=6.6 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}\right)$
  1. $8.34 \times 10^{-27} \mathrm{~kg}$
  2. $0.0083 \mathrm{~kg}$
  3. $0.83 \times 10^{-27} \mathrm{~kg}$
  4. $0.8 \times 10^{-28} \mathrm{~kg}$

Solution

Precision $= \pm 1 \%$ $\therefore$ Accuracy $=0.01$ $\begin{aligned} & \Delta v=3 \times 10^8 \times \frac{2}{100}=6 \times 10^6 \mathrm{~m} / \mathrm{s} \\ & \Delta x=1.05 \times 10^{-13} \mathrm{~m}\end{aligned}$ We know that, $\Delta x \cdot \Delta v=\frac{h}{4 \pi m}$ $\Rightarrow \quad m=\frac{6.626 \times 10^{-34}}{4 \pi \times 1.05 \times 10^{-13} \times 6 \times 10^6}$ $\begin{aligned} & =0.08 \times 10^{-27} \mathrm{~kg} \\ & =0.8 \times 10^{-28} \mathrm{~kg}\end{aligned}$

Asked in: AP EAMCET 2021 (23 Aug Shift 2)

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