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

I am really confused on how to go about 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 $m = 0.100~\mathrm{fg}$ (where a femtogram, $\mathrm{fg}$, is $10^{−15}~\mathrm{g}$) and is swimming at a velocity of $v = 9.00~\mathrm{{\mu m/s}}$, with an uncertainty in the velocity of $9.00\%$. E. coli bacterial cells are around $1~\mathrm{\mu m}$ ($10^{−6}~\mathrm{m}$) 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 $[m] = 0.100~\mathrm{fg} = 0.1 \times 10^{-15}~\mathrm{g} = 10^{-16}~\mathrm{g}$
• Velocity of bacteria $[v] = 2~\mathrm{m/s}$
• 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 '16 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. – user23061 Mar 28 '16 at 13:26