Question

An electron travelling at 3.7 x 105 m/s has an uncertainty in its velocity of 1.88 x 105 m/s. What is the uncertainty in its position (in pm)? (Just put the number, don't use scientific notation and decimal point in the final answer) Avogadro's number = 6.022 x 1023, h = 6.626 x 10-34 J.s, c = 3.00 x 108 m/s, me= 9.109 x 10-31 kg

Answer 1

Answer:

The uncertainty in its position is 308 pm

Explanation:

Given;

velocity of electron, v = 3.7 x 10⁵ m/s

uncertainty in velocity of the electron, Δv =   1.88 x 10⁵ m/s

Planck's constant, h = 6.626 x 10⁻³⁴ J.s

mass of electron, m = 9.109 x 10⁻³¹ kg

Using Heisenberg's uncertainty principle, we will determine the uncertainty in the position of the electron.

Δx(mΔv) ≥ h/4π

where;

Δx is uncertainty in position

m is mass of the electron

Δv is uncertainty in velocity of the electron

h is Planck's constant

Δx $= \frac{h}{4\pi (mv) } = \frac{6.626*10^{-34}}{4\pi (9.109*1.88*10^5)} = 3.08*10^{-10} \ m$

Δx = 308 x 10⁻¹² m

Δx = 308 pm

Therefore,  the uncertainty in its position is 308 pm