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
1s2s2p
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#.

Related questions
Impact of this question
1065 views around the world
You can reuse this answer
Creative Commons License