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:

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