Answer :
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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