How to apply the Heisenberg Uncertainty Principle in calculations on the movement of bacteria?

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.

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

$\newcommand{\upmu}{{\large\unicode[Times]{x3BC}}}%Remove this line when upright greek characters are implemented by mathjax$ (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 $h = 6.62607004\times 10^{-34}~\mathrm{m^2\,kg\,s^{-1}}$.

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

• 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.