Question

Radioactive isotopes often occur together in mixtures. Suppose a 220g sample contains ¹³¹Ba, with a half-life of 12 days, and ⁴⁷Ca, with a half-life of 4.5 days.

If there are initially twice as many calcium atoms as there are barium atoms, what will be the ratio of calcium atoms to barium atoms 2.10 weeks later?

N_Ca/N_Ba = ?

Answer 1

Total Mass M₀ = 220 g

Let initial Ba mass = M₀₁

Let the initial no. of Ba atoms : N₀₁

Initial no. of Calcium atoms= twice the no of Ba atoms : N₀₂ = 2N₀₁

Atomic mass 131 Ba = 130.91 u

A = Avogadro Number = 6.023 * 10²³ atoms in 130.9 gm of Ba

No. of atoms of Ba: N₀₁: (130.91/A)*M₀₁

No. of atoms of Ca: N₀₂ = 2*M₀₁* (130.91/A)

Ratio: N_02 / N_01 = 2

Total no. of atoms initially = 2N₀₁ + N₀₁ = 3N₀₁

Radioactive decay equation gives the no. of nuclei present at any time t:

Thus for Ba:

And Ca:

In above equations t is in days.

We divide N)2(t) by N_1(t)

This becomes:

Since N_02 = 2 * N_01

or,

Now substitute the value of t in days and get the required ratio.

Since it is confusing from your question whether t = 2 weeks or 10 weeks or something else,

I will do it for both 2 and 10 weeks .

1. If t = 2 weeks = 14 days:

2. If t = 10 weeks, t = 70 days:

If t = something else just substitute the value (in days) in above equation to get the required answer.