Respuesta :
Answer: 2.1 %
Explanation:
The radius of the Argon atom, r = 71 pm = 7.1 × 10 ⁻¹¹ m
Average orbital speed of electrons, v = 3.9 × 10⁷ m/s
From uncertainty principle:
Δx m Δv ≥ h/4π
mass of electron, m = 9.1 ×10⁻³¹ kg
Δx = radius of the argon atom = 7.1 × 10 ⁻¹¹ m
[tex]\Rightarrow \Delta v = \frac {6.626 \times 10^{-34} m^2kg/s}{4\times 3.14 \times 7.1 \times 10^{-11} m \times 9.1 \times 10^{-31} kg}[/tex]
[tex]\Delta v = 8.2 \times 10^5 m/s[/tex]
Percentage uncertainty:
[tex]\frac{\Delta v}{v} \times 100\% = \frac {8.2 \times 10^5 m/s}{3.9 \times 10^7 m/s} \times 100 \%= 2.1 \%[/tex]
The least possible uncertainty in a measurement of the speed of the electron in an atom of argon is 1.0%
Given the data in the question;
Radius of argon; [tex]r = 71 pm[/tex]
Diameter of argon; [tex]\delta x = 2r = 2*71 pm = 142pm = 1.42*10^{-10}m[/tex]
Average orbital speed; [tex]v_{orb} = 3.9 * 10^7 m/s[/tex]
According to Heisenberg uncertainty principle or indeterminacy principle:
[tex]\delta x\ \delta p \geq \frac{h}{4\pi }\\\\\delta x\ m\delta v \geq \frac{h}{4\pi }[/tex]
Where:
- [tex]\delta x[/tex] is the uncertainty in position
- [tex]\delta p[/tex] is the uncertainty of momentum
- [tex]h[/tex] is the Planck's constant ( [tex]6.626 * 10^{-34} kg.m^2/s[/tex] )
- [tex]\pi[/tex] is pi
- [tex]\delta v[/tex] is uncertainty speed
- m is the mass of electron ( [tex]9.1*10^{-31}kg[/tex] )
Hence
[tex]\delta v = \frac{h}{4\pi \ *\ \delta x\ *\ m}[/tex]
We substitute in our values
[tex]\delta v = \frac{6.626*10^{-34}kg.m^2/s}{4\pi \ *\ (1.42*10^{-10}m)\ *\ (9.1*10^{-31}kg)}\\\\\delta v = \frac{6.626*10^{-34}kg.m^2/s}{1.6238*10^{-39}kg.m}\\\\\delta v = 408055.179 m/s[/tex]
Now, Uncertainty as a percentage of the average speed;
= ( uncertainty in speed / average speed ) × 100%
[tex]= \frac{408055.179m/s}{3.9*10^7m/s}\ *\ 100[/tex]
[tex]=1.0[/tex]%
Therefore, the least possible uncertainty in a measurement of the speed of the electron in an atom of argon is 1.0%
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