In this problem, I know the formula, but I get confused with all the steps and how to use what I got, to get to what I need.

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of $m = \pu{0.100 fg}$ (where a femtogram, $\pu{fg}$, is $\pu{10^{−15} g}$) and is swimming at a velocity of $v = \pu{9.00\mu m/s}$, with an uncertainty in the velocity of $9.00\%$. E. coli bacterial cells are around $\pu{1 \mu m}$ ($\pu{10^{−6} m}$) in length. What is the uncertainty of the position of the bacterium?

I used the formula $$\Delta x \Delta p \geq \hslash/2$$ I plugged in these numbers:

  • Mass of bacteria $(m) = \pu{0.100 fg} = \pu{0.1 \times 10^{-15} g} = \pu{10^{-16} g}$
  • Velocity of bacteria $(v) = \pu{2 m/s}$
  • Uncertainty in velocity = 9.00

I am getting this answer $1.67\times10^{-9}$, and it is wrong. I don't know what I am doing wrong.

  • 3
    $\begingroup$ Well why don't you tell us what you know and then we might be able to help. $\endgroup$
    – bon
    Mar 27, 2016 at 9:13

1 Answer 1


$\newcommand{\upmu}{{\large\unicode[Times]{x3BC}}}%Remove this line when upright greek characters are implemented by mathjax$ (Hover over the boxes to see the results.) The velocity of the bacteria is $\mathbf{v} = 9~\mathrm{\upmu m/s} = 9\times10^{-6}~\mathrm{m/s}$, with an uncertainty of $9.00\%$ you can calculate the absolute uncertainty in the velocity $\Delta \mathbf{v}$ as the difference of the upper bound $(\mathbf{v} + 0.09\mathbf{v})$ and the lower bound $(\mathbf{v} - 0.09\mathbf{v})$.

$$\Delta \mathbf{v} = (\mathbf{v} + 0.09\mathbf{v}) - (\mathbf{v} - 0.09\mathbf{v}) = 0.18\mathbf{v} = 1.62\times10^{-6}~\mathrm{m\,s^{-1}}$$

The momentum is defined as $$\mathbf{p} = m\mathbf{v},$$ hence the absolute uncertainty in the momentum is $$\Delta\mathbf{p} = m\Delta\mathbf{v}.$$ The mass is $m = 0.100~\mathrm{fg} = 1.00\times10^{-19}~\mathrm{kg}$.

$$\Delta\mathbf{p} = m\Delta\mathbf{v} = 1.62\times10^{-25}~\mathrm{kg\,m\,s^{-1}}$$

Now you can use this to calculate the uncertainty in position $\Delta\mathbf{x}$. $$\Delta\mathbf{x}\Delta\mathbf{p} \geq \frac{h}{2}$$ Planck's constant is $h = 6.62607004\times 10^{-34}~\mathrm{m^2\,kg\,s^{-1}}$.

$$\Delta\mathbf{x} \geq \frac{h}{2\Delta\mathbf{p}}\\\Delta\mathbf{x} \geq 2.07\times10^{-9}~\mathrm{m}$$

Since you have not included any of the steps how you arrived at your result, I cannot help you finding your mistake.

  • $\begingroup$ I think we can argue about the fact that the uncertainty is on the magnitude of the velocity or on the velocity itself. If the result is still different from OP's corrections, the uncertainty can be divided by 2 and the result should correspond to the solutions. $\endgroup$
    – user23061
    Mar 28, 2016 at 10:26
  • $\begingroup$ @R.M. I cannot make sense of what you have written. Also what corrections are you talking about? $\endgroup$ Mar 28, 2016 at 10:45
  • $\begingroup$ The OP states that its result is wrong, so I am assuming that he is comparing it with some solutions. Concerning you answer, I interpreted the statement "with an uncertainty in the velocity of 9.00%" as $\Delta v = 0.09 |\vec{v}|$, which is half of your uncertainty. I whink that the two interpretations are equally correct. $\endgroup$
    – user23061
    Mar 28, 2016 at 13:26

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