Answer:
Velocity of electron along x direction is 57.9 m/s
Explanation:
The uncertainty in x coordinate of electron, Δx = 0.200 mm = 0.2 x 10⁻³ m
Let vₓ be the x component of electrons velocity.
The uncertainty in x component of electrons momentum is:
Δpₓ = mΔvₓ
Here m is mass of the electron.
The uncertainty in velocity x component is 1% i.e. 0.01.
So, the above equation can be written as :
Δpₓ = 0.01mvₓ ....(1)
The minimum uncertainty principle is:
[tex]\Delta x\Delta p_{x} = \frac{h}{2\pi }[/tex] ....(2)
Here h is Planck's constant.
From equation (1) and (2),
[tex]\Delta x\times0.01m v_{x} = \frac{h}{2\pi }[/tex]
Substitute 0.2 x 10⁻³ m for Δx, 9.1 x 10⁻³¹ kg for m and 6.626 x 10⁻³⁴ m²kg/s in the above equation.
[tex]0.2\times10^{-3} \times0.01\times9.1\times10^{-31}\times v_{x} = \frac{6.626\times10^{-34} }{2\pi }[/tex]
vₓ = 57.9 m/s
Given the uncertainty in the measurement of position, the velocity of the electron is 57.7 m/s.
Using the relation;
ΔxΔpx≥ℏ
where Δx is the uncertainty in the x coordinate of a particle, Δpx is the particle's uncertainty in the x component of momentum
and ℏ=h/2π
Recall that h = 6.6 × 10^-34 Js
Also, Δpx = Δmv
where, m = mass, v = velocity
Hence;
Δx Δmv = h/2π
Recall that m = mass of the electron = 9.11 × 10^-31 Kg
Δx = 0.200 mm or 2 × 10^-4 m
Δv = h/2πΔx Δm
Δv = [tex]\frac{6.6* 10^-34 Js}{2 * 3.14 * (2 * 10^-4 m) * (9.11 *10^-31 Kg)}[/tex]
Δv = 0.577
Since the uncertainty in the measurement is 1.00 %, v = 57.7 m/s
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