Question

Suppose the velocity of an electron in an atom is known to an accuracy of 2.0×103m/s (reasonably accurate compared with orbital velocities). What is the electron’s minimum uncertainty in position, and how does this compare with the approximate 0.1-nm size of the atom?

Answer 1

Answer:

289.714 times bigger

Explanation:

$\Delta x$ = Uncertainty in position

$\Delta p$ = Uncertainty in momentum = $\Delta v m$

$\Delta v$ = Uncertainty in velocity = $2\times 10^3\ m/s$

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

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

From the Heisenberg uncertainty principle we have

$\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta x\Delta v m=\dfrac{h}{4\pi}\\\Rightarrow \Delta x=\dfrac{h}{4\pi\Delta v m}\\\Rightarrow \Delta x=\dfrac{6.626\times 10^{-34}}{4\pi \times 2\times 10^3\times 9.1\times 10^{-31}}\\\Rightarrow \Delta x=2.89714\times 10^{-8}\ m$

Comparing with 0.1 nm size atom

$\dfrac{\Delta x}{x}=\dfrac{2.89714\times 10^{-8}}{0.1\times 10^{-9}}\\\Rightarrow \dfrac{\Delta x}{x}=289.714$

So, the electron’s minimum uncertainty in position is 289.714 times bigger than an atom of size 0.1 nm