Q35 / 180 · AIIMS 2018 (26 May) · Chemistry

The position of both an electron and helium atom is known within $1.0 \mathrm{~nm}$. The momentum of the electron is known within $5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$. The minimum uncertainty in the measurement of the momentum of the helium atom is

  • A
    $7.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$
  • Correct answer
    B
    $5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$
  • C
    $8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$
  • D
    $6.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

Solution

The Heisenberg uncertainty principle, $\Delta x \times \Delta p \geq \frac{h}{4 \pi}$, where $\Delta x=$ Uncertainty in position, $\Delta p=$ Uncertainty in momentum and $h / 4 \pi=$ constant. As $\Delta x$ is same for electron and helium and $h / 4 \pi$ is a constant, therefore minimum uncertainty in the measurement of the momentum of the helium atom will be same as that of an electron which is $5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$.

Hence, the correct answer is option: 'B'.

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