Question

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.500 fg (where a femtogram, fg, is 10−15g) and is swimming at a velocity of v = 7.00 μm/s , with an uncertainty in the velocity of4.00 %.. E. coli bacterial cells are around 1 μm ( 10−6 m) in length. The student is supposed to observe the bacterium and make a drawing. However, the student, having just learned about the Heisenberg uncertainty principle in physics class, complains that she cannot make the drawing. She claims that the uncertainty of the bacterium's position is greater than the microscope's viewing field, and the bacterium is thus impossible to locate.

What is the uncertainty of the position of the bacterium?

Express your answer with the appropriate units.

Answer 1

The uncertainty of the position of the bacterium =
`3.76*10^(-10) \ m`

Step-by-step explanation:

Given that:

mass (m) =0.500 fg

To kilogram; we have:

`m = (0.500)/(10^(18))`

`m = 5*10^(-19) \ kg`

Velocity (v) = 7.00 μm/s

To meter/seconds (m/s);

Velocity (v) =
`7*10^(-6) \ m/s`

Uncertainty of the velocity is given as 4% = 0.04

Then; multiplying the velocity of the bacterium; we have:

`7*10^(-6) \ m/s`
`*0.04`

`= 2.8*10^(-7)`

To determine the uncertainty in the momentum;we multiply the uncertainty in the velocity by the mass:

`m* \delta y = (2.8*10^(-7) )(5*10^(-19) ) \\ \\ m* \delta y = 1.4*10^(-25)`

Now; according to Heisenberg's Uncertainty Principle;

`\delta x* m * \delta y \geq (h)/(4 \pi ) \\ \\ \\ \delta x = (h)/(4 \pi*(m*\delta y))`

where;

`\delta x` = uncertainty in the position

`\delta y =` uncertainty in the velocity

h = Planck's Constant

`\delta x = (h)/(4 \pi*(m*\delta y))`

`\delta x = (6.626*10^(-34))/(4 \pi*(1.4*10^(-25)))`

`\delta x = (6.626*10^(-34))/(1.76*10^(-10))`

`\delta x =3.76*10^(-10) \ m`

Thus, the uncertainty of the position of the bacterium =
`3.76*10^(-10) \ m`