Technicium-99m has a half-life of 6.00 hours? plot the decay of 800. g of technicium-99m for 5 half-lives

I don't understand how to graph or answer the question

1 Answer
May 10, 2018

For g:
800e^(-xln(2)/6),x in[0,30]
graph{800e^(-xln(2)/6) [0, 30, -100, 1000]}

or

For kg:
0.8e^(-xln(2)/6),x in[0,30]
graph{0.8e^(-xln(2)/6) [0, 30, -0.1, 1]}

Explanation:

The exponential decay equation for a substance is:
N=N_0e^(-lambdat), where:

  • N = number of particles present (though mass can be used too)
  • N_0 = number of particles at the start
  • lambda = decay constant (ln(2)/t_(1/2)) (s^-1)
  • t = time (s)

To make things easier, we will keep the half life in terms of hours, while plotting time in hours. It doesn't really matter what unit you use as long as t and t_(1/2) are both using the same units of time, in this case it is hours.

So, N_0=800g (or 0.8kg)
t_(1/2)=6.00 "hours"
t=30 "hours" (since 5 half-lives would be 30 hours)

So, plot a graph of y=800e^(-xln(2)/6),x in[0,30] if you are using grams or y=0.8e^(-xln(2)/6),x in[0,30] if you are using kilograms. the graph would be mass (g or kg) against time (hours).

If you made to draw it, then plot several values of y for different values of x.