Answered

(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 :

skyluke89

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

b) The kinetic energy of the ion is [tex]5.9\cdot 10^{-15} eV[/tex]

Explanation:

a)

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

[tex]\Delta p \Delta x \geq \frac{h}{4\pi}[/tex]

where

[tex]\Delta p[/tex] is the uncertainty on the momentum, which can be rewritten as

[tex]\Delta p = m \Delta v[/tex], where

[tex]m[/tex] is the mass

[tex]\Delta v[/tex] is the uncertainty on the velocity

[tex]\Delta x[/tex] is the uncertainty on the position

[tex]h=6.63\cdot 10^{-34} Js[/tex] is the Planck constant

In this problem, we have

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

[tex]m = 5.86\cdot 10^{-26}kg[/tex] is the mass of the ion

Re-arranging the equation and solving for [tex]\Delta v[/tex], we find:

[tex]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[/tex]

b)

The kinetic energy of the ion is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is the mass of the ion

v is its velocity

Here we have:

[tex]m = 5.86\cdot 10^{-26}kg[/tex] is the mass of the ion

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

Substituting,

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

Converting into electronvolts,

[tex]K=\frac{9.5\cdot 10^{-34}}{1.6\cdot 10^{-19}}=5.9\cdot 10^{-15} eV[/tex]

which is approximately [tex]10^{15}[/tex] smaller than the typical molecular binding energy of 5 eV.

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The minimum uncertainty in velocity and kinetic energy in eV is mathematically given as

  • dv>1.80 * 10^{-4}m/s
  • K.E=5.9* 10^{-15} eV

What is the minimum uncertainty in velocity, and kinetic energy in eV?

Question Parameter(s):

Generally, the equation for Heisenberg's uncertainty principle  is mathematically given as

[tex]m dx dv \geq \frac{h}{4\pi}\[/tex]

Therefore

[tex]dv \geq \frac{h}{4\pi m dx}\\dv > \frac{6.63*10^{-34}}{4\pi (5.86*t 10^{-26})(5.0*10^{-6})}[/tex]

dv>1.80 * 10^{-4}m/s

In conclusion, The kinetic energy of the ion

K,E=0.5mv^2

Therefore

[tex]K.E=\frac{1}{2}(5.86\cdot 10^{-26})(1.80\cdot 10^{-4})^2\\[/tex]

K.E=9.5*t 10^{-34} J

Therefore

[tex]K=\frac{9.5* 10^{-34}}{1.6* 10^{-19}}[/tex]

K.E=5.9* 10^{-15} eV

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