The x coordinate of an electron is measured with an uncertainty of 0.200 mm . What is vx, the x component of the electron’s velocity, if the

Question

The x coordinate of an electron is measured with an uncertainty of 0.200 mm . What is vx, the x component of the electron’s velocity, if the minimum percentage uncertainty in a simultaneous measurement of vx is 1.00 % ? Use the following expression for the uncertainty principle:

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=h2π, where h is Planck’s constant.

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1 year 2021-08-31T21:35:01+00:00 1 Answers 121 views 0

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:

$$\Delta x\Delta p_{x} = \frac{h}{2\pi }$$    ….(2)

Here h is Planck’s constant.

From equation (1) and (2),

$$\Delta x\times0.01m v_{x} = \frac{h}{2\pi }$$

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.

$$0.2\times10^{-3} \times0.01\times9.1\times10^{-31}\times v_{x} = \frac{6.626\times10^{-34} }{2\pi }$$

vₓ = 57.9 m/s