Modifying an equation to plot a straight line.

In summary: You are correct. The log10 of the actual quantity is plotted on the y-axis. This is 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.
  • #1
thatguythere
91
0

Homework Statement


Show all the steps required to modify the equation so that the plot yields a straight line.
N/N0=[itex]e\ [/itex]-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.
 
Last edited:
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  • #2
Well, do you get a straight line when you plot it? If you do, you're done.
 
  • #3
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.
 
  • #4
thatguythere said:

Homework Statement


Show all the steps required to modify the equation so that the plot yields a straight line.
N/N0=[itex]e\ [/itex]-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
 
  • #5
thatguythere said:
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.
 
  • #6
tms said:
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.

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

ln(N/N0)=-ux

ehild
Exactly.
 
  • #7
Redbelly98 said:
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.
 
  • #8
thatguythere said:
[
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:
Puck1956.jpg

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)
 

1. How do I modify an equation to plot a straight line?

To modify an equation to plot a straight line, you need to rewrite it in the form y = mx + b, where m represents the slope of the line and b represents the y-intercept. This format is known as the slope-intercept form and is commonly used to plot straight lines.

2. What does the slope of a line represent?

The slope of a line represents the rate of change between the two variables being plotted. It tells you how much the dependent variable (y) changes for every unit increase in the independent variable (x). In other words, it is the steepness of the line.

3. How do I find the slope of a line from an equation?

To find the slope of a line from an equation, you can use the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are any two points on the line. This formula is also known as the slope formula.

4. Can an equation with a negative slope still plot a straight line?

Yes, an equation with a negative slope can still plot a straight line. A negative slope simply means that the line is decreasing from left to right, as opposed to increasing from left to right with a positive slope. As long as the equation is in the form y = mx + b, it will always plot a straight line.

5. How do I determine the y-intercept from an equation?

The y-intercept can be determined by looking at the value of b in the slope-intercept form y = mx + b. It represents the point where the line intersects the y-axis. For example, if the equation is y = 2x + 3, the y-intercept is 3, and the line will cross the y-axis at the point (0,3).

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