Question

The half-life of 235U, an alpha emitter, is 7.1×108 yr. Part A Calculate the number of alpha particles emitted by 4.4 mg of this nuclide in 2 minutes. Express your answer using two significant figures.

Answer 1

First, use Avogadro's Number to figure out how many U235 atoms are in 4.4 mg

1 mole of U235 contains 6.02×10^23 atoms

No = (0.0044 grams)(6.02*10^23 atoms/(235 grams)) = 1.05×10^19 atoms

By definition, the number of U235 atoms remaining after time "t" is:

N(t) = (No)e^(-t/)

Where = (half-life)/ln(2)
= (4.1*10^8 yr)(3.16*10^7 sec/yr)/ln(2)
= 1.86*10^16 seconds

The number that have DECAYED after time t is:

K(t) = No - N(t)
= No - (No)e^(-t/)
= No(1 - e^(-t/))

The RATE of decay (# of decays per second) at time "t" is:

R(t) = dK/dt
= (No/)e^(-t/)

At t=0 (i.e. right now), the decay rate is:

Ro = (No/)e^(-0/)
= No/

Since the half life is very, very large compared to 2 minutes, we can safely say that the decay rate will be constant (will equal No/) for the entire 2-minute duration. That means:

Number of decays in 2 min = Ro * 2 min
= (No/)(120 sec)

= (1.05*10^19 atoms/1.86*10^16 seconds ) (120 sec)

= 67,742 atoms

So, the number of alpha particles emitted by 4.4 mg of this nuclide in 2 minutes= 67,742