Q33PE - College Physics (Urone)

Question

Approximately how many atoms thick is a cell membrane, assuming all atoms there average about twice the size of a hydrogen atom? 

Step-by-Step Solution

Answer

The thickness of a cell membrane is equal to the combined thickness of \(50{\rm{ atoms}}\).

Step 1: Thickness in terms of atoms:

The thickness in terms of atoms is defined by using the following formula.

\(t = \frac{{{d_{{\rm{membrane}}}}}}{{2 \times {d_{H{\rm{ atom}}}}}}\) 

Step 2: Diameter of Hydrogen:

The diameter of a hydrogen atom, \({d_{H{\rm{ atom}}}} = {10^{ - 10}}{\rm{ m}}\).

The thickness of the cell membrane, \({d_{{\rm{membrane}}}} = {10^{ - 8}}{\rm{ m}}\).

Step 3: Solving for known data:

Write the equation for the thickness in terms of atom as below.

\(t = \frac{{{d_{{\rm{membrane}}}}}}{{2 \times {d_{H{\rm{ atom}}}}}}\)

Substitute \({10^{ - 8}}{\rm{ m}}\) for \({d_{{\rm{membrane}}}}\) and \({10^{ - 10}}{\rm{ m}}\) for \({d_{H{\rm{ atom}}}}\) in the above equation.

\(\begin{array}{c}t = \frac{{{{10}^{ - 8}}{\rm{ m}}}}{{2 \times {{10}^{ - 10}}{\rm{ m}}}}\\ = \frac{{{{10}^2}}}{2}\\ = \frac{{100}}{2}\\ = 50{\rm{ atoms}}\end{array}\)

Step 4: Conclusion:

Hence, the thickness of a cell membrane is equal to the combined thickness of \(50{\rm{ atoms}}\).