The position of both, an electron and a helium atom is known within $1.0 \mathrm{~nm}$. Further the momentum…

The position of both, an electron and a helium atom is known within $1.0 \mathrm{~nm}$. Further the momentum of the electron is known within $5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~ms}^{-1}$. The minimum uncertainty in the measurement of the momentum of the helium atom is
  1. $50 \mathrm{~kg} \mathrm{~ms}^{-1}$
  2. $80 \mathrm{~kg} \mathrm{~ms}^{-1}$
  3. $8.0 \times 10^{-26} \mathrm{~kg} \mathrm{~ms}^{-1}$
  4. $5.0 \times 10^{-26} \mathrm{~kg} \mathrm{~ms}^{-}$

Solution

According to Heisenberg uncertainty principle,
$\Delta x \times \Delta p=\frac{h}{4 \pi}$ (which is constant).
As $\Delta x$ for electron and helium atom is same, thus momentum of electron and helium will also be same, therefore the momentum of helium atom is equal to $5 \times 10^{-26} \mathrm{~kg} \mathrm{~ms}^{-1}$.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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