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Physics Ncert Solutions Class 12th Physics Class 12th Nuclei

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

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Payal Gupta | Contributor-Level 10

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13.25 Half life of ${}_{15}^{32}P {, T}_{1 / 2}$ = 14.3 days

Half life of ${}_{15}^{33}P, T_{1 / 2}$ = 25.3days and ${}_{15}^{33}P$ decay is 10% of the total amount of decay

The source has initially 10 % of ${}_{15}^{33}P$ nucleus and 90% of ${}_{15}^{32}P$ nucleus

Suppose after t days, the source has 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{33}P$ nucleus

Initially:

Number of ${}_{15}^{33}P$ nucleus = N

Number of ${}_{15}^{32}P$ nucleus = 9N

Finally:

Number of ${}_{15}^{33}P$ nucleus = 9N'

Number of ${}_{15}^{32}P$ nucleus = N'

For ${}_{15}^{32}P$ nucleus, the number ratio, $\frac{N'}{9 N}$ = ${(\frac{1}{2})}^{t / T_{1 / 2}}$

N' = 9N ${(2)}^{- t / 14.3}$ …………….(1)

For ${}_{15}^{33}P$ nucleus, the number ratio, $\frac{9 N'}{N}$ = ${(\frac{1}{2})}^{t / T_{1 / 2}}$

9N' = N ${(2)}^{- t / 25.3}$ …………….(2)

On dividing equation (1) by equation (2), we get:

$\frac{1}{9}$ = 9 $* 2^{(\frac{t}{25.3} - \frac{t}{14.3})}$

$\frac{1}{81}$ = $2^{(\frac{- 11 t}{25.3 * 14.3})}$

Taking l

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13.25 Half life of ${}_{15}^{32}P {, T}_{1 / 2}$ = 14.3 days

Half life of ${}_{15}^{33}P, T_{1 / 2}$ = 25.3days and ${}_{15}^{33}P$ decay is 10% of the total amount of decay

The source has initially 10 % of ${}_{15}^{33}P$ nucleus and 90% of ${}_{15}^{32}P$ nucleus

Suppose after t days, the source has 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{33}P$ nucleus

Initially:

Number of ${}_{15}^{33}P$ nucleus = N

Number of ${}_{15}^{32}P$ nucleus = 9N

Finally:

Number of ${}_{15}^{33}P$ nucleus = 9N'

Number of ${}_{15}^{32}P$ nucleus = N'

For ${}_{15}^{32}P$ nucleus, the number ratio, $\frac{N'}{9 N}$ = ${(\frac{1}{2})}^{t / T_{1 / 2}}$

N' = 9N ${(2)}^{- t / 14.3}$ …………….(1)

For ${}_{15}^{33}P$ nucleus, the number ratio, $\frac{9 N'}{N}$ = ${(\frac{1}{2})}^{t / T_{1 / 2}}$

9N' = N ${(2)}^{- t / 25.3}$ …………….(2)

On dividing equation (1) by equation (2), we get:

$\frac{1}{9}$ = 9 $* 2^{(\frac{t}{25.3} - \frac{t}{14.3})}$

$\frac{1}{81}$ = $2^{(\frac{- 11 t}{25.3 * 14.3})}$

Taking log on both sides

log 1 – log 81 = $\frac{- 11 t}{25.3 * 14.3}$ log2

0 – 1.908 = ( $\frac{- 11 t}{361.79}$ ) $* 0.301$

t = 208.5 days

Hence, it will take about 208.5 days for 90% decay of $P_{15}^{33}$

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13.25 Half life of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>32</mn> </mrow> </mrow> </mmultiscripts> <msub> <mrow> <mrow> <mo>,</mo> <mi> </mi> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msub> </math> = 14.3 daysHalf life of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> <mo>,</mo> <mi> </mi> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msub> </math>  = 25.3days and <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> </math>  decay is 10% of the total amount of decayThe source has initially 10 % of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> </math> nucleus and 90% of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>32</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleusSuppose after t days, the source has 10% of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>32</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleus and 90% of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleusInitially:Number of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> </math> nucleus = NNumber of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>32</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleus = 9N Finally:Number of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> </math> nucleus = 9N’Number of <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>32</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleus = N’For <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>32</mn> </mrow> </mrow> </mmultiscripts> </math> nucleus, the number ratio, <math> <mfrac> <mrow> <mrow> <mi>N</mi> <mi>'</mi> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> <mi>N</mi> </mrow> </mrow> </mfrac> </math>  = <math> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mi>t</mi> <mo>/</mo> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </msup> </math> N’ = 9N <math> <msup> <mrow> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mi>t</mi> <mo>/</mo> <mn>14.3</mn> </mrow> </mrow> </msup> </math> …………….(1) For <math> <mmultiscripts> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mprescripts></mprescripts> <mrow> <mrow> <mn>15</mn> </mrow> </mrow> <mrow> <mrow> <mn>33</mn> </mrow> </mrow> </mmultiscripts> </math>  nucleus, the number ratio, <math> <mfrac> <mrow> <mrow> <mn>9</mn> <mi>N</mi> <mi>'</mi> </mrow> </mrow> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> </mfrac> </math> = <math> <msup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </mrow> </mrow> <mrow> <mrow> <mi>t</mi> <mo>/</mo> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msub> </mrow> </mrow> </msup> </math> 9N’ = N <math> <msup> <mrow> <mrow> <mo>(</mo> <mn>2</mn> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mi>t</mi> <mo>/</mo> <mn>25.3</mn> </mrow> </mrow> </msup> </math>  …………….(2) On dividing equation (1) by equation (2), we get: <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </mfrac> </math> = 9 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mo>(</mo> <mfrac> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>25.3</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>14.3</mn> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </msup> </math> <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>81</mn> </mrow> </mrow> </mfrac> </math>  = <math> <msup> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mo>(</mo> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>11</mn> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>25.3</mn> <mo>×</mo> <mn>14.3</mn> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </msup> </math> Taking log on both sideslog 1 – log 81 = <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>11</mn> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>25.3</mn> <mo>×</mo> <mn>14.3</mn> </mrow> </mrow> </mfrac> </math> log20 – 1.908 = ( <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>11</mn> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>361.79</mn> </mrow> </mrow> </mfrac> </math> ) <math> <mo>×</mo> <mn>0.301</mn> </math> t = 208.5 daysHence, it will take about 208.5 days for 90% decay of <math><mmultiscripts><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mn>15</mn></mrow></mrow><mrow><mrow><mn>33</mn></mrow></mrow></mmultiscripts></math>

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

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

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