Chapter 42: Q. 54 (page 1237)

Radioactive isotopes often occur together in mixtures. Suppose a $100 g$ sample contains $131 B a$ , with a half-life of $12 d a y s$ , and $47 C a$ , with a half-life of localid="1650485393662" $4.5 d a y s$ . 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 localid="1650485404787" $2.5$ weeks later?

Short Answer

Expert verified

Ratio of calcium to barium atoms is, $\frac{N_{C a}}{N_{B a}} = 0.373$

Step by step solution

Given information

We are given that,

Half life of Ba is, ${(t_{\frac{1}{2}})}_{B a} = 12 d a y s$

Half life of Ca is,localid="1650485477257" ${(t_{\frac{1}{2}})}_{C a} = 4.5 d a y s$

Simplify

Let time constant for calcium is, $τ_{C a}$

And time constant for barium is, $τ_{B a}$

Time constant,

localid="1650485605612" $τ_{C a} = \frac{{(t_{\frac{1}{2}})}_{C a}}{\ln (2)} = \frac{4.5 d a y s}{\ln (2)} = 6.492 d a y s$

And,

localid="1650485623266" $τ_{B a} = \frac{{(t_{\frac{1}{2}})}_{B a}}{\ln (2)} = \frac{12 d a y s}{\ln (2)} = 17.30 d a y s$

Now, ratio is,

localid="1650485643779" $\frac{N_{C a}}{N_{B a}} = \frac{{(N_{a})}_{C a}}{{(N_{a})}_{B a}} \times \frac{e^{\frac{- t}{τ_{C a}}}}{e^{\frac{- t}{τ_{B a}}}} = \frac{2 \times 0.0675}{0.3639} = 0.373$

Here, $N_{C a}$ is final no. of atoms of calcium

And $N_{B a}$ is final no. of atoms of barium

And ${(N_{α})}_{C a}$ is initial no. of atoms of calcium

And ${(N_{α})}_{B a}$ is initial no. of atoms of barium

And $t$ is time after which ratio of atoms is calculated

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Simplify

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