Question

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 1.80 fg (where a femtogram, fg, is 10−15g) and is swimming at a velocity of v = 6.00 μm/s , with an uncertainty in the velocity of 4.00 % . 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.

Answer 1

Given mass of bacteria, m = 1.80 fg * (10^-15 g / 1 fg) * (1Kg / 1000 g) = 1.80*10^-18 Kg

Velocity of bacteria, v = 6.00*10^-6 m/s

=> Momentum of the bacteria, P = m*v = 1.80*10^-18 Kg * 6.00*10^-6 m/s = 1.08*10^-23 Kg.m/s

Given uncertainity in velocity, $\Delta$v = 4.00%

Hence uncertainity in momentum, $\Delta$P is also 4.00 %

=> $\Delta$P = P * (4.00/100) = 1.08*10^-23 Kg.m/s * (4.00 / 100) = 4.32*10^-25 Kg.m/s

Applying Heisenberg uncertainty principle:

$\Delta$X*$\Delta$P $\geqslant$ h/2

71

where $\Delta$X = uncertainity in position

and h = Planck's constant = 6.626*10^-34 m².kg / s

=> $\Delta$X $\geqslant$ h/(2
71
*$\Delta$P)

=> $\Delta$X $\geqslant$ 6.626*10^-34 m².kg.s^-1 / (2*3.14*4.32*10^-25 Kg.m.s^-1)

=> $\Delta$X $\geqslant$ 2.44*10^-10 m

=> $\Delta$X $\geqslant$ 2.44*10^-10 m * (1 mn / 10^-9 m) = 0.244 nm

Since a microscope's viewing field is larger than 0.244 nm, there will be no problem in viewing the bacteria and hence no problem in drawing.

Hence the student is wrong.