Question

The position of a Helium atom is restricted to 0.40nm. The uncertainty in its momentum is

Answer 1

Answer:

$\\\tt \approx 1.3 \times 10^{-25} kg ms^{-1}$

Explanation:

Using:

$\\\tt \Delta x . \Delta p\geq \dfrac{h}{4 \pi}$

Here delta x represents position restriction and delta p represents momentum restriction:

Using

Δx $\\\tt = 4 \times 10^{-10}m$

We get the uncertainty in momentum as :

$\\\tt (4 \times 10^{-10} . \Delta p) \geq \dfrac{h}{4 \pi}\\\implies \Delta p \geq \dfrac{h}{4 \pi \times 4 \times 10^{-10} } \\\\\implies \Delta p \approx 1.3 \times 10^{-25} kg ms^{-1}$