Question

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

Answer 1

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`.