The speed of the electron is measured to be 5 ｘ10³ m s⁻¹ to an accuracy of 0.003 %. Find the uncertainty in determining the position of this electron. (Answer: 4.84 ｘ 10⁻³ m )



Given:

Speed of electron = v = 5 ｘ10³ m s⁻¹

% accuracy in the speed of electron = 0.003% 

∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹
∴ Mass of the electron = m = 9.11 ｘ 10⁻³¹ kg

To Find:

Uncertainty in position = Δx = ?



Solution:

According to Heisenberg's uncertainty principle, 

Δx ΔP ≈ h

or

Δx ≈ `\frac {h}{ΔP}` --------------(1)

Where h is Plank's constant and ΔP is the uncertainty in momentum.

to find uncertainty in momentum we need to find uncertainty in velocity, So

uncertainty in velocity =  Δv = 0.003% of 5 ｘ10³ m s⁻¹
=`\frac {0.003}{100}` ｘ 5 ｘ10³ m s⁻¹ = 0.15 m s⁻¹

Now ΔP = mΔv = 9.11 ｘ 10⁻³¹ kg ｘ 0.15 m s⁻¹ = 1.3665 ｘ 10⁻³¹ kg m s⁻¹

Now putting the corresponding values in equation (1)

Δx ≈ `\frac {6.626 ｘ 10⁻³⁴ m² kg s⁻¹}{1.367 ｘ 10⁻³¹ kg m s⁻¹}`

Δx ≈ 4.847 ｘ 10⁻³ m 

Δx ≈ 4.847 mm --------------------Ans.


The uncertainty in the position of an electron is 4.852 millimeters.


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