Question

(a) If the position of a chlorine ion in a membrane is measured to an accuracy of 5.00 µm, what is its minimum uncertainty in velocity, given its mass is 5.86 ✕ 10-26 kg?
m/s
(b) If the ion has this velocity, what is its kinetic energy in eV? (Compare this with typical molecular binding energies of about 5 eV.)
eV

Answer 1

a) The uncertainty on the velocity is $1.80 \cdot 10^{-4}m/s$

b) The kinetic energy of the ion is $5.9\cdot 10^{-15} eV$

Explanation:

a)

We can solve this part by using the Heinsenberg's uncertainty principle, which states that:

$\Delta p \Delta x \geq \frac{h}{4\pi}$

where

$\Delta p$ is the uncertainty on the momentum, which can be rewritten as

$\Delta p = m \Delta v$, where

$m$ is the mass

$\Delta v$ is the uncertainty on the velocity

$\Delta x$ is the uncertainty on the position

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

In this problem, we have

$\Delta x = 5.00 \mu m= 5.0\cdot 10^{-6} m$ is the uncertainty on the position

$m = 5.86\cdot 10^{-26}kg$ is the mass of the ion

Re-arranging the equation and solving for $\Delta v$, we find:

$m\Delta x \Delta v \geq \frac{h}{4\pi}\\\Delta v \geq \frac{h}{4\pi m \Delta x}=\frac{6.63\cdot 10^{-34}}{4\pi (5.86\cdot 10^{-26})(5.0\cdot 10^{-6})}=1.80 \cdot 10^{-4}m/s$

b)

The kinetic energy of the ion is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the ion

v is its velocity

Here we have:

$m = 5.86\cdot 10^{-26}kg$ is the mass of the ion

$v=1.80\cdot 10^{-4} m/s$ is the velocity

Substituting,

$K=\frac{1}{2}(5.86\cdot 10^{-26})(1.80\cdot 10^{-4})^2=9.5\cdot 10^{-34} J$

Converting into electronvolts,

$K=\frac{9.5\cdot 10^{-34}}{1.6\cdot 10^{-19}}=5.9\cdot 10^{-15} eV$

which is approximately $10^{15}$ smaller than the typical molecular binding energy of 5 eV.

Learn more about kinetic energy:

brainly.com/question/6536722

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