Modifying an equation to plot a straight line.

[thatguythere]

Homework Statement

Show all the steps required to modify the equation so that the plot yields a straight line.
N/N₀=e\^-ux

This equation demonstrates the fraction radiation absorbed by a material, where "N₀" is the number of incident photons from the radioactive source without any absorbed introduced, "N" is the number of transmitted photons, "u" is the absorption coefficient of the absorber (units m^-1) and x is the thickness of the absorber.

Homework Equations

The Attempt at a Solution

Subsequently I am instructed to graph lnN/N₀ vs "x". Therefore I am assuming N/N₀ will be my "y" value in the equation of the line, whereas "x" (the thickness) will be my "x" value. I tried something along these lines, but have no idea if I am even in the ballpark.

N/N₀=e^-ux
lnN/N₀=-ux

That seems to give me lnN/N₀ as "y", -u as "m", x as "x" and 0 as "b". Please give me some thoughts. Thank you.

 

[tms]

Well, do you get a straight line when you plot it? If you do, you're done.

 

[thatguythere]

The problem is that I have no data to plug into the equation and verify it because this is a question that I need to have answered before we do the experiment.

 

[ehild]

Homework Statement

Show all the steps required to modify the equation so that the plot yields a straight line.
N/N₀=e\^-ux

This equation demonstrates the fraction radiation absorbed by a material, where "N₀" is the number of incident photons from the radioactive source without any absorbed introduced, "N" is the number of transmitted photons, "u" is the absorption coefficient of the absorber (units m^-1) and x is the thickness of the absorber.

Homework Equations

The Attempt at a Solution

Subsequently I am instructed to graph lnN/N₀ vs "x". Therefore I am assuming N/N₀ will be my "y" value in the equation of the line, whereas "x" (the thickness) will be my "x" value. I tried something along these lines, but have no idea if I am even in the ballpark.

N/N₀=e^-ux
lnN/N₀=-ux

That seems to give me lnN/N₀ as "y", -u as "m", x as "x" and 0 as "b". Please give me some thoughts. Thank you.

It is all right, but you have to write N/N₀ in parentheses.

ln(N/N₀)=-ux

ehild

 

[tms]

The problem is that I have no data to plug into the equation and verify it because this is a question that I need to have answered before we do the experiment.

Just let x run from 0 to the total thickness.

 

[Redbelly98]

Just let x run from 0 to the total thickness.

That's not a bad suggestion. Just plug into Excel or other graphing program, and look at the graph. You'd need to make numbers up for u and N₀, but it would show whether you can expect a straight line.

It is all right, but you have to write N/N₀ in parentheses.

ln(N/N₀)=-ux

ehild

Exactly.

 

[tms]

That's not a bad suggestion. Just plug into Excel or other graphing program, and look at the graph. You'd need to make numbers up for u and N₀, but it would show whether you can expect a straight line.

In my universe, all constants equal 1.

 

[CompuChip]

[
That seems to give me lnN/N₀ as "y", -u as "m", x as "x" and 0 as "b". Please give me some thoughts. Thank you.

That sounds good. This is what is called a logarithmic plot (or a log-linear plot, to indicate that one of the axes is logarithmic and the other is just linear). An exponential function such as the one you have for N, will become a straight line in such a plot. This is particularly useful since for a relatively small range of x, the y-values usually go through a wide range of values and taking the logarithm keeps it all a bit more easily viewable.

Note that you can also use the property ln(a/b) = ln(a) - ln(b) to rewrite the equation to
ln(N) = - u x + ln(N₀)
in which case you will get a log-linear plot for N and your starting value "b" will be the initial value N₀ (although on your logarithmic y-axis, you will actually plot ln(N₀).

[edit]Here is another example of a log-plot:

Note how equal distances on the y-axis correspond to multiplications instead of additions, in other words, they've plotted the log₁₀ of the actual quantity.

(Googled it from http://www.eyephysics.com/tdf/models.htm)