The accompanying graph illustrates the decay of 88 Mo, which decay via positron emission.?

What is the half life for the decay?

What is the half life for the decay?
You can see from the graph that the mass has decreased by half after 7 minutes. That’s the half-life.

—-
What is the rate constant for the decay?
Decay rate constant k = [ln(2) / half-life] = [ln(2) / 7]

—-
What fraction of the original sample remains after 12 mins?
N(t) = N₀ * e^(-kt)

N(t) = amount at time t
N₀ = original amount
k = decay constant
t = time

N(12) = 1.0 * e*[(-ln(2) * 12) / 7]

≈ 0.30 ………….. to 2 significant figures

As a fraction, 3/10 of the original amount remains after 12 minutes
(which matches the graph)

—-
What is the product of the decay process?

(88/42) Mo ⟶ (88/41) Nb + (0/1) e + (0/0) v_e

The (0/1) e on the product side is a positron.

The (0/0) v_e on the product side is an electron neutrino.

Leave a Comment