The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days). Concept introduction: According to first order of kinetics: ln ( N N 0 ) = − λ t Here, N and N o are final and initial concentrations, λ is decay constant and t is time. t 1 / 2 ( h a l f − t i m e ) = ln 2 λ Here, t 1 / 2 is half-life time.

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Answer 1

Question

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

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Answer 2

Question

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

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Answer 3

Question

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

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Answer 4

Question

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

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Answer 5

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Answer 6

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Answer 7

Chapter 32, Problem 32.3QAP

Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=−λt

Here, N and N_o are final and initial concentrations, λ is decay constant and t is time.

t1/2(half−time)=ln2λ

Here, t1/2 is half-life time.

Answer 8

Expert Solution & Answer

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=−λt

Where,

t1/2 = half time

λ = decay constant

N and N₀ = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements	t1/2		t1/2 hr	λ hr-1
Iron-59	44.51 days		1068.24 hr	0.00065
Titanium-45	3.078 hr		3.078 hr	0.22519
Calcium-47	4.536 days		108.864 hr	0.00637
Phosphorous-33	25.3 days		607.2 hr	0.00114
For t=24 hr	t hr	24	t= 48 hr	t hr	48
	ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	2.021×10−5
Calcium-47	-0.1528	0.085829		-0.3056	0.7366662
Phosphorous-33	-0.0274	0.97297		-0.0548	0.9466799
	t =72 hr	t hr	72	t = 96 hr	t hr	96
		ln(N/N0)	(N/N0)		ln(N/N0)	(N/N0)
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	9.1×10−8		-21.618626	4.1×10−10
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*−1CellC11=EXP(B11)CellE11=E5*F9*−1CellF11=EXP(E11)CellC18=E5*D16*−1CellD18=EXP(C18)CellF18=E5*G16*−1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Answer 12

Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element	t hr ln(N/N0)	24 (N/N0)	t hr ln(N/N0)	48 (N/N0)	t hr ln(N/N0)	72 (N/N0)	t hr ln(N/N0)	96 (N/N0)
Iron-59	0.0156	0.98455	-0.0311	0.9693343	-0.0467	0.95436	-0.0622914	0.93961
Titanium-45	-5.4047	0.0045	-10.809	2.021×10−5	-16.214	9.1×10−8	-21.618626	4.1×10−10
Calcium-47	-0.1528	0.085829	-0.3056	2.021×10−5	-0.4584	0.63228	-0.6112409	0.54268
Phosphorous-33	-0.0274	0.97297	-0.0548	0.9466799	-0.0822	0.9211	-0.1095885	0.8962

Answer 13

Ch. 32 - Identify X in each of the following nuclear...Ch. 32 - Prob. 32.2QAP Ch. 32 - Prob. 32.3QAP Ch. 32 - Prob. 32.4QAP Ch. 32 - Prob. 32.5QAP Ch. 32 - Prob. 32.6QAP Ch. 32 - Prob. 32.7QAP Ch. 32 - Prob. 32.8QAP Ch. 32 - Prob. 32.9QAP Ch. 32 - Prob. 32.10QAP

Concept explainers

Answer to Problem 32.3QAP

Explanation of Solution

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Chapter 32 Solutions