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:

$\ln \left(\frac{N}{N_0}\right)=-\lambda t$

Here, N and N_o are final and initial concentrations, $\lambda$ is decay constant and t is time.

$t_{1/2}\left(half-time\right)=\ln \frac{2}{\lambda }$

Here, $t_{1/2}$ is half-life time.

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 $\ln \left(N/N_0\right)$	24 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	48 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	72 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	96 $\left(N/N_0\right)$
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	$2.021\times {10}^{-5}$	-16.214	$9.1\times {10}^{-8}$	-21.618626	$4.1\times {10}^{-10}$
Calcium-47	-0.1528	0.085829	-0.3056	$2.021\times {10}^{-5}$	-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-

$t_{1/2}=\ln \frac{2}{\lambda }$

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

$\ln \left(\frac{N}{N_0}\right)=-\lambda t$

Where,

$t_{1/2}$ = half time

$\lambda$ = 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	$t_{1/2}$		$t_{1/2}$ hr	$\lambda$ 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
	$\ln \left(N/N_0\right)$	$\left(N/N_0\right)$		$\ln \left(N/N_0\right)$	$\left(N/N_0\right)$
Iron-59	0.0156	0.98455		-0.0311	0.9693343
Titanium-45	-5.4047	0.0045		-10.809	$2.021\times {10}^{-5}$
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
		$\ln \left(N/N_0\right)$	$\left(N/N_0\right)$		$\ln \left(N/N_0\right)$	$\left(N/N_0\right)$
		-0.0467	0.95436		-0.0622914	0.93961
		-16.214	$9.1\times {10}^{-8}$		-21.618626	$4.1\times {10}^{-10}$
		-0.4584	0.63228		-0.6112409	0.54268
		-0.0822	0.9211		-0.1095885	0.8962

Spreadsheet documentation of the above table is −

$\begin{array}{l}CellE5=LN\left(2\right)/D5\hfill \\ CellB11=E5*C9*-1\hfill \\ CellC11=EXP\left(B11\right)\hfill \\ CellE11=E5*F9*-1\hfill \\ CellF11=EXP\left(E11\right)\hfill \\ CellC18=E5*D16*-1\hfill \\ CellD18=EXP\left(C18\right)\hfill \\ CellF18=E5*G16*-1\hfill \\ CellG18=EXP\left(F18\right)\hfill \end{array}$

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

Radioactive element	t hr $\ln \left(N/N_0\right)$	24 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	48 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	72 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	96 $\left(N/N_0\right)$
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	$2.021\times {10}^{-5}$	-16.214	$9.1\times {10}^{-8}$	-21.618626	$4.1\times {10}^{-10}$
Calcium-47	-0.1528	0.085829	-0.3056	$2.021\times {10}^{-5}$	-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 $\ln \left(N/N_0\right)$	24 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	48 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	72 $\left(N/N_0\right)$	t hr $\ln \left(N/N_0\right)$	96 $\left(N/N_0\right)$
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	$2.021\times {10}^{-5}$	-16.214	$9.1\times {10}^{-8}$	-21.618626	$4.1\times {10}^{-10}$
Calcium-47	-0.1528	0.085829	-0.3056	$2.021\times {10}^{-5}$	-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