A biology student must use a microscope to track a single bacterium traveling along the length of a cell. The bacterium is $1.20$ $f"g"$ and travels at a speed of $1.00 xx 10^(-6) (pm 5.00%) "m/s"$ [ . . . ] ?

Md Ariful Islam

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Md Ariful Islam

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The student is not correct. The uncertainty is at least $8.88 xx 10^(-4)$ $mu"m"$, but that is over $1000$ times smaller than the minimum viewing window needed to see the whole bacterium.

Alright, well, we can tell that this is about the , because obviously the question talks about it.

$barul(|" "stackrel(" ")DeltaxDeltap >= ℏ//2" "|)$

  • $Deltax$ is the uncertainty in the position.
  • $Deltap = mDeltav$ is the uncertainty in the momentum.
  • $ℏ = h//2pi$ is the reduced .
  • $m$ is the mass of the bacterium in $"kg"$.

We assume the bacterium is restricted to one dimension, and that that dimension is along the cell length.

Noting the conversions needed to get from $"g" -> "kg"$, $mu "m" -> "m"$, and $f"g" -> "g"$, the uncertainty in the position is then:

$color(blue)(Deltax) >= (ℏ//2)/(mDeltav)$

$= overbrace((6.626 xx 10^(-34) cancel"kg"cdot"m"^(cancel(2))"/"cancel"s" // 4 pi))^(ℏ//2)/(underbrace(1.20 xx 10^(-15) "g" xx cancel"1 kg"/(1000 cancel"g"))_(m) xx underbrace(0.0500 xx 1.00 xx 10^(-6) cancel"m/s")_(Deltav))$

$>= color(blue)(8.88 xx 10^(-10))$ $color(blue)("m")$ or $8.88 xx 10^(-4)$ $mu"m"$.

So the student is not correct.

The uncertainty is at most $1126.12$ times smaller than the minimum required viewing window to which the microscope should be adjusted, and the image should be clear... microscopes for bacterium are fairly standard fare...

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