A source contains two phosphorous radio nuclides ${}_{15}^{32}P (T_{1 / 2} = 14.3 d)$ and ${}_{15}^{33}P (T_{1 / 2} = 25.3 d) .$ Initially, 10% of the decays come from ${}_{15}^{33}P .$ How long one must wait until 90% do so?

Solution

We know that rate of disintegration is,
$- \frac{dN}{dt} \propto N .$

So, clearly the initial ratio of the phosphorous radio nuclides, ${}_{15}^{33}P and {}_{15}^{32}P$ is 1:9.

We have to find the time at which the ratio is 9:1.

Initially, if the amount of ${}_{15}^{33}P$ is x, then the amount of ${}_{15}^{32}P$ is 9x.

Finally, if the amount of ${}_{15}^{33}P$ is 9y, the amount of ${}_{15}^{32}P$ is y.

Now, from the formula,
$\frac{N}{N_{o}} = {(\frac{1}{2})}^{n} = {(\frac{1}{2})}^{\frac{t}{T}} = 2^{\frac{- t}{T}}$

Therefore,
Using,    $N = \frac{N_{0}}{2^{t / T}}$

$\Rightarrow$ $9 y = \frac{x}{2^{t / 25.3}} y = \frac{9 x}{2^{t / 14.3}}$

On dividing,    $9 = \frac{x}{2^{t / 25.3}} \times \frac{2^{t / 14.3}}{9 x}$

$\Rightarrow$    $81 = 2^{\frac{t}{14.3} - \frac{t}{25.3}}$
$\Rightarrow$    $81 = 2^{\frac{11 t}{361.79}}$

$\Rightarrow$ $\log_{10} 81 = \frac{11 t}{361.79} \log_{10} 2 = \frac{11 \times 0.3010 t}{361.79}$

$\Rightarrow$    $9.15 \times 10^{- 3} t = 1.91$

i.e., $t = \frac{1.91 \times 1000}{9.15} d = 208.7 days$

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Wave Optics

A source contains two phosphorous radio nuclides ${}_{15}^{32}P (T_{1 / 2} = 14.3 d)$ and ${}_{15}^{33}P (T_{1 / 2} = 25.3 d) .$ Initially, 10% of the decays come from ${}_{15}^{33}P .$ How long one must wait until 90% do so?

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Wave Optics

A source contains two phosphorous radio nuclides P1532 (T1/2 = 14.3 d) and P1533 (T1/2 = 25.3 d). Initially, 10% of the decays come from P1533. How long one must wait until 90% do so?

Some More Questions From Wave Optics Chapter

Mock Test Series

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A source contains two phosphorous radio nuclides ${}_{15}^{32}P (T_{1 / 2} = 14.3 d)$ and ${}_{15}^{33}P (T_{1 / 2} = 25.3 d) .$ Initially, 10% of the decays come from ${}_{15}^{33}P .$ How long one must wait until 90% do so?