Question

The radioisotope sodium-24 (24Na11), half life 15 h, is used to measure the
flow rate of salt water. By irradiation of Stable 23Na11 with neutrons, suppose
that we produce 5 micrograms of isotope. How much do we have at the end
of 24h?

Answer 1

Answer –

We are given, half-life t ½ = 15 h , initial amount Ni = 5 micrograms

Nt = ? time = 24 h

The irradiation of Stable 23Na11 with neutrons is first order reaction.

First we need to calculate the decay constant from the given half life

We know,

Decay constant = 0.693 / half life

                          = 0.693 / 15 h

                           = 0.0462 h^-1

Now we need to calculate the Nt, means the remaining amount of sodium-24 (24Na11) after 24 h

ln Nt/ Ni = -k*t

ln Nt / 5 micrograms = - 0.0462 h^-1 * 24 h

ln Nt / 5 micrograms = -1.108

taking antiln from both side

Nt/ 5 micrograms = 0.329

Nt = 0.329 * 5 micrograms

     = 1.65 micrograms

So, 1.65 micrograms of isotope have at the end of 24h.