Question

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

• A. $7.0 \times 10^{-26} kg\, ms ^{-1}$
• B. $5.0 \times 10^{-26} kg \,ms ^{-1}$
• C. $8 \cdot 0 \times 10^{-26} \,kg\, ms ^{-1}$
• D. $6 \cdot 0 \times 10^{-26} kg \,ms ^{-1}$

Answer 1

The Correct Option is B

The Heisenberg uncertainty principle,

$\Delta x \times \Delta p \leq \frac{h}{4 \pi}$, where $\Delta x=$ Uncertainty in position,

$\Delta p=$ Uncertainty in momentum and $\frac{h}{4 \pi}=$ constant.

As $\Delta x$ is same for electron and helium and $\frac{h}{4 \pi}$ is a constant, therefore minimum uncertainty in the measurement of the momentum of the helium atom will be same as tlat of an electron which is $5.0 \times 10^{-26} \,kg /\, ms ^{-1}$