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Physics Question on momentum

The position of both an electron and helium atom is known within 1.0nm1.0 nm. The momentum of the electron is known within 5.0×10−26kgms−15.0 \times 10^{-26} kg ms ^{-1}. The minimum uncertainty in the measurement of the momentum of the helium atom is

A

7.0×10−26kg ms−17.0 \times 10^{-26} kg\, ms ^{-1}

B

5.0×10−26kg ms−15.0 \times 10^{-26} kg \,ms ^{-1}

C

8⋅0×10−26 kg ms−18 \cdot 0 \times 10^{-26} \,kg\, ms ^{-1}

D

6⋅0×10−26kg ms−16 \cdot 0 \times 10^{-26} kg \,ms ^{-1}

Answer

5.0×10−26kg ms−15.0 \times 10^{-26} kg \,ms ^{-1}

Explanation

Solution

The Heisenberg uncertainty principle,

Δx×Δp≤h4π\Delta x \times \Delta p \leq \frac{h}{4 \pi}, where Δx=\Delta x= Uncertainty in position,

Δp=\Delta p= Uncertainty in momentum and h4π=\frac{h}{4 \pi}= constant.

As Δx\Delta x is same for electron and helium and h4π\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 5.0×10−26 kg/ ms−15.0 \times 10^{-26} \,kg /\, ms ^{-1}