Question

German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δv)Δx≥h4πmΔvwhere h is Planck's constant and m is the mass of the object.The mass of an electron is 9.11×10−31 kg.What is the uncertainty in the position of an electron moving at 3.00×106 m/s with an uncertainty of Δv=0.01×106 m/s?Δ

Answer 1

Answer:

5.79*10⁻⁹ m is the uncertainty in the position.

Explanation:

Heisenberg's uncertainty principle assumes that it is not possible to know exactly all the data regarding the behavior of particles. In other words, at the subatomic level, it is impossible to know at the same moment where a particle is, how it moves and what its speed is.

So, Heisenberg's Uncertainty Principle gives a relationship between the standard deviation of an object's position and its momentum.

Δp*Δx= h/(4π)

where

• Δp the standard deviation of the object's momentum,
• Δx the standard deviation of the object's position,
• h=6.63*10⁻³⁴ J.s is the Planck's constant.

By definition, the momentum of the electron equals the product of its mass and velocity. So, being the mass constant, you can said:

Δp= m*Δv

Replacing in the expresion of the Heisenberg's Uncertainty Principle:

m*Δv*Δx= h/(4π)

Then you know:

• m=9.11*10⁻³¹ kg
• Δv=0.01*10⁶ m/s

• h=6.63*10⁻³⁴ J.s= 6.63*10⁻³⁴ (N*m)*s=6.63*10⁻³⁴ [(kg*m*s⁻²)*m]*s= 6.63*10⁻³⁴ kg*m²*s⁻¹

Replacing:

*Δx=6.63*10⁻³⁴ kg*m²*s⁻¹/(4π)

Taking π=3.14 and solving:

Δx$=\frac{6.63*10^{-34} kg*m^{2} *s^{-1} }{4*3.14*9.11*10^{-31} kg*0.01*10^{6} m/s}$

Δx=5.79*10⁻⁹ m

5.79*10⁻⁹ m is the uncertainty in the position.