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