Help with these problems?

Question

can anybody show me the complete solutions for these problems?

1.If approximately 600 of a certain bacteria can fit across your Low Power field of vision, what is the approximate size of one bacterium????

2.If 30 objects fit across a Low Power field of view whose field diameter is 3000 micros,what is the approximate size of each object???

tnx a lot 4 those who can show me the solution!!

Answer 1

The area of a 3000 micron circular field is:

A = (pi(3000)^2)/4 micron^2

1. If the type of bacteria is globular ( coccus or spherical like Thiomargarita Nimibienses), then the area occupied by a bacterium including the space between four bacteria in one layer is D^2, where D is the diameter of a bacterium, Thus;

D^2 =( pi (3000)^2)/(4 x600)

D = 108.54 microns

If the bacteria is bacillus shaped type (like Epulopiscum Fishelsoni), then the area occupied by a bacterium including the space between  four bacteria is d x l, where d is the diameter of the bacterium and l is the length of a bacterium.

Thus dl = (pi(3000)^2)/(4 x600)

Since the average diameter of a bacillus bacterium is about 80 microns, the length of a bacterium in the colony occupying the field is;

l = 147.262 microns.

2. Again, if the objects are round, then the area occupied by one object including the area between four objects is D^2, where D is the diameter of one object. Thus;

D^2 = (pi x (3000)^2)/(4x30)

D = 485.41 microns

Answer 2

not enough info..try to ask again...

Answer 3

1. not enough information.

2. 3000/30 = 100 micros (micrometers?)