Question

If there is 10umol of the radioactive isotope ³²P (half-life 14 days) at t=0, how much ³²P will remain at (a). 7 days (b) 14 days (c) 21 days and (d) 70 days?

I already know the answers, I have the student companion guide to my biochemistry textbook. It does not tell me how to work out the problem though. If anyone could help it would be appreciated!

The answers are (a) 7mmol (b) 5umol (c) 3.5mmol and (d) 0.3umol

Answer 1

Answer – Given, initial amount = 10 umol, half-life = 14 days,

First we need to calculate the decay constant from half life

We know, radioactive decay is the first order kinetic energy

So, k = 0.693 / t ½

        = 0.693 / 14 days

       = 0.0495 day^-1

Time, t = 7 days

We know the formula for first order reaction

ln Nt / No = - k* t

ln Nt / 10 umol = - 0.0495 day^-1 * 7 day

ln Nt / 10 umol = -0.346

We taking the antiln from both side

Nt / 10 umol = 0.707

Nt = 10 umol * 0.707

     = 7.0 umol

Now when t = 14 days

ln Nt / 10 umol = - 0.0495 day^-1 * 14 day

ln Nt / 10 umol = -0.693

We taking the antiln from both side

Nt / 10 umol = 0.500

Nt = 10 umol * 0.50

Nt = 5.0 umol

t = 21 days

ln Nt / 10 umol = - 0.0495 day^-1 * 21 day

ln Nt / 10 umol = -1.04

We taking the antiln from both side

Nt / 10 umol = 0.35

Nt = 10 umol * 0.35

Nt = 3.5 umol

When t = 70 days

ln Nt / 10 umol = - 0.0495 day^-1 * 70 day

ln Nt / 10 umol = -3.46

We taking the antiln from both side

Nt / 10 umol = 0.0313

Nt = 10 umol * 0.0313

Nt = 0.3 umol