Question

An atom of silver has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of silver. Write your answer as a percentage of the average speed, and round it to significant digits.

Answer 1

Answer:

Percentage of uncertainty in average speed of an electron is 0.1756%.

Explanation:

Using Heisenberg uncertainty principle:

$\Deltax\times \Delta p\geq \frac{h}{4\pi }$

$\Delta p=m\times \Delta v$

$\Deltax\times m\times \Delta v\geq \frac{h}{4\pi }$

Δx = Uncertainty in position

Δp = Uncertainty in momentum

Δv = Uncertainty in average speed

h = Planck's constant = $6.626\times 10^{-34} kg m^2/s$

m = mass of electron =  $9.1\times 10^{-31} kg$

We have

Δx = 2 × 165 pm = 330 pm = $3.3\times 10^{-10} m$

$1 pm = 10^{-12} m$

Average orbital speed of electron = v = $=1.0\times 10^8 m/s$

$3.3\times 10^{-10} 9.1\times 10^{-31} kg \times \Delta v\geq \frac{6.626\times 10^{-34} kg m^2/s}{4\pi }$

$\Delta v\geq \frac{6.626\times 10^{-34} kg m^2/s}{4\pi \times 3.3\times 10^{-10}\times 9.1\times 10^{-31} kg}$

$\Delta v\geq 1.756\times 10^5 m/s$

Percentage of uncertainty in average speed:

$=\frac{\Delta v}{v}\times 100$

$=\frac{1.756\times 10^5 m/s}{1.0\times 10^8 m/s}\times 100=0.1756\%$