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

Half life of ^ 32 _ 15 P ,T_ 1/2 =14.3 days Half life of ^ 33 _ 15 P ,T_ 1/2 =25.3 days ^ 33 _ 15 P nucleus decay is 10% of the total amount of decay. The source has initially 10% of ^ 33 _ 15 P nucleus and 90% of ^ 32 _ 15 P nucleus. suppose after t days, the source has 10% of ^ 32 _ 15 P nucleus and 90% of ^ 33 _ 15 P nucleus. Initially: Number of ^ 33 _ 15 P nucleus = N Number of ^ 32 _ 15 P nucleus = 9 N Finally: Number of ^ 33 _ 15 P nucleus = 9 N' Number of ^ 32 _ 15 P nucIeus = N' For ^ 32 _ 15 P nucleus, we can write the number ratio as: N' 9N = (1 2 )^ t T_v2 N'=9N(2)^ -t 14.3 (1) For ^ 33 _ 15 P , we can write the number ratio as: 9N' N = (1 2 )^ t T_v2 9N'=N(2)^ -t 25.3 (2) On dividing equation (1) by equation (2), we get: 1 9 =9 2 ( t 25.3 - t 14.3 ) 1 81 =2- ( 11t 25.3 14.2 ) 1- 81= -11t 25.3 14.3 2 -11t 25.3 14.3 =0-1.908 0.301 t=25.3 14.3 1.908 11 0.301 208.5 days Hence, it will take about 208.5 days for 90% decay of _ 15 P^ 33 .

A source contains two phosphorous radio nuclides ^ 32 _ 15 P (T_ …

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

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Answer

Half life of $^{32}_{15}\text{P },\text{T}_{1/2}=14.3\text{ days}$
Half life of $^{33}_{15}\text{P },\text{T}_{1/2}=25.3\text{ days}$
$^{33}_{15}\text{P}$ nucleus decay is 10% of the total amount of decay.
The source has initially 10% of $^{33}_{15}\text{P}$ nucleus and 90% of $^{32}_{15}\text{P}$ nucleus.
suppose after t days, the source has 10% of $^{32}_{15}\text{P}$ nucleus and 90% of $^{33}_{15}\text{P}$ nucleus.
Initially:
Number of $^{33}_{15}\text{P}$ nucleus = N
Number of $^{32}_{15}\text{P}$ nucleus = 9 N
Finally:
Number of $^{33}_{15}\text{P}$ nucleus = 9 N'
Number of $^{32}_{15}\text{P}$ nucIeus = N'
For $^{32}_{15}\text{P}$ nucleus, we can write the number ratio as:
$\frac{\text{N}'}{9\text{N}}=\Big(\frac{1}{2}\Big)^{\frac{\text{t}}{\text{T}_\text{v2}}}$
$\text{N}'=9\text{N}(2)^{\frac{-\text{t}}{14.3}}\ \dots(1)$
For $^{33}_{15}\text{P },$ we can write the number ratio as:
$\frac{9\text{N}'}{\text{N}}=\Big(\frac{1}{2}\Big)^{\frac{\text{t}}{\text{T}_\text{v2}}}$
$9\text{N}'=\text{N}(2)^{\frac{-\text{t}}{25.3}}\ \dots(2)$
On dividing equation (1) by equation (2), we get:
$\frac{1}{9}=9\times2\Big(\frac{\text{t}}{25.3}-\frac{\text{t}}{14.3}\Big)$
$\frac{1}{81}=2-\Big(\frac{11\text{t}}{25.3\times14.2}\Big)$
$\log1-\log81=\frac{-11\text{t}}{25.3\times14.3}\log2$
$\frac{-11\text{t}}{25.3\times14.3}=\frac{0-1.908}{0.301}$
$\text{t}=\frac{25.3\times14.3\times1.908}{11\times0.301}\approx208.5\text{ days}$
Hence, it will take about 208.5 days for 90% decay of $_{15}\text{P}^{33}.$

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