Question

The half-life of 235U, an alpha emitter, is 7.1×108 yr,Calculate the number of alpha particles emitted by 4.1 mg of this nuclide in 2 minutes

Answer 1

First, use Avogadro's Number to figure out how many U235 atoms are in 4.1 mg

1 mole of U235 contains 6.02×10^23 atoms

No = (0.0041 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/λ)

Where λ = (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/λ)
= No(1 - e^(-t/λ))

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

R(t) = dK/dt
= (No/λ)e^(-t/λ)

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

Ro = (No/λ)e^(-0/λ)
= No/λ

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/λ) for the entire 2-minute duration. That means:

Number of decays in 2 min = Ro × 2 min
= (No/λ)(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 4.1 mg of this nuclide in 2 minutes= 67,742