how many disintegrations have occurred after 11 minutes?

Question

For 1.4 mole of the radioactive isotope molybdenum-99, half-life = 67 hours, how many disintegrations have occurred after 11 minutes

Answer 1

The radioactive decay equation is:

N(t) = N(0)*exp(-L*t)

where N(0) is the quantity of radioactive material present initially

N(t) is the quantity of radioactive material remaining at time t

L is the radioactive decay constant = ln(2)/halflife

The number of decays that have happened between times t=0 and t=t is then N(0) - N(t), so

# decays = N(0)*(1-exp(-L*t))


In this case, L = ln(2)/67 hr = 1.724*10^-1 min^-1

N(0) = 1.4 moles

# decays =1.4 moles * (1 - exp(-1.724*10^-1 min^-1 * 11 min))

# decays = 2.653*10^-3 mol

Multiplying by Avogadro's number to get the number of actual decays:

# decays = 2.653*10^-3 * 6.022*10^23 atoms/mol

# decays = 1.598*10^21 atoms decayed

Answer 2

The number of disintegrations, D, will be equal to N*(1 - 0.5^(t/T)), where N is the initial number of particles, t is the elapsed time, and T is the half-life. N is 1.4 moles, which is 1.4*6.02 x 10^23 = 8.4308 x 10^23. T can be rewritten as (67 hours)(60 min/hr) = 4020 min. So we have D = (8.4308 x 10^23)*(1 - 0.5^(11/4020)), which you can calculate yourself.