Collection of Solved Problems in Physics

Flying electron

Task number: 4402

• Hint 1

  We know that, using the principle of uncertainty, with the specification of the electron’s motion comes the price of lost accuracy in measuring another quantity. What is the quantity (or quantities) in question?

• Solution of hint 1

  With the use of the principle of uncertainty, we know that the specification of the electron’s motion comes with the price of accuracy. Thus, the inaccuracy of velocity, momentum and kinetic energy measurement for the particle rises (these three variables can be derived from each other).

  Do you know any particular version of such a principle of uncertainty?

• Solution of hint 1

  The principle of uncertainty for the location and momentum has the form (in 1D) $$\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2} .$$

  As the uncertainty of the momentum consider in this case the absolute uncertainty of determination of the momentum.

• Hint 2

  What is the relative uncertainty in the measurement? How does the relative uncertainty propagate from one variable to another? What if the one variable is the product of two other which carry some uncertainty themselves?

• Solution of hint 2

  The relative uncertainty is a dimensionless quantity (often in percentage). We can calculate it from the absolute uncertainty of the measurement divided by the measured value. We can denote it with the letter η.

  If the variable C is the product of variables A and B, we can use the formula η(C) = η(A) + η(B).

• Notation

  me = 9.11·10−31 kg	Rest mass of the electron
  v = 1000 km·s−1	Electron’s velocity
  Δv = 1 km·s−1	Absolute uncertainty of velocity
  Δx = ? (m)	Uncertainty of location

• Solution

  We need to know the uncertainty Δp to estimate Δx from the principle of uncertainty $$\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2}$$ . As uncertainty of the momentum, we consider the absolute uncertainty of determination of the momentum. Since the ball’s momentum is the product of its mass and velocity p = m_e·v. If we consider the mass of the electron as a constant the uncertainty of the momentum is given only by uncertainty of the velocity. Therefore we can use the formula η(p) = η(v) to calculate the uncertainty of these variables. We have also the formula between absolute and relative uncertainty Δp = η(p)·p. With the use of the principle of uncertainty we can calculate Δx $$\Delta x \,\cdot\, \Delta p \ge \frac{\hbar}{2}$$ and we obtain

  $$\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\Delta p} ,$$ $$\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\eta(p)\cdot p} ,$$ $$\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\eta(v)\cdot mv} ,$$ $$\Delta x \ge \frac{\hbar}{2} \,\frac{1}{\frac{\Delta v}{v}\cdot mv} ,$$ $$\Delta x \ge \frac{1.055\,\cdot\,10^{-34}}{2\,\cdot\,\frac{1}{1000}\,\cdot\, 9.11\,\cdot\, 10^{-25}}\ \mbox{m} ,$$ $$\Delta x \ge 5.8\,\cdot\,10^{-8}\ \mbox{m}\,\dot{=}\,60\ \mbox{nm} .$$

  Out of principle we cannot determine the location of the mentioned electron with the accuracy higher than tens of nanometres, which is two orders of magnitude higher than the characteristic dimensions of an atom. From the electron’s perspective, this uncertainty is quite important.

• Discussion

  Lorentz factor

  $$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\left( \frac{1}{300}\right)^2}}=\sqrt{\frac{90\,000}{89\,999}}\,\dot{=}\, 1.000 006 .$$

  If the momentum of our electron is in the order of 10⁻²⁴ kg·m·s⁻¹ we could see the relativist effects in order of 10⁻³⁰ kg·m·s⁻¹ so we can neglect it.

• Answer

  The minimum uncertainty of the electron’s location is approximately 60 nm.