need answer thanks! only answer if you know please​!

Answer:

a) [tex]k \approx 0.0602\,\frac{1}{min}[/tex] ([tex]k\approx 6.02\,\frac{\%}{min}[/tex]), b) [tex]k \approx 0.0648\,\frac{1}{min}[/tex] ([tex]k \approx 6.48\,\frac{\%}{min}[/tex]), c) [tex]t_{1/2} \approx 60.274\,days[/tex], d) [tex]t_{1/2}\approx 4.305\,min[/tex], e) [tex]k \approx 0.3\,\frac{1}{min}[/tex] ([tex]k\approx 30\,\frac{\%}{min}[/tex]), f) [tex]k \approx 0.0120\,\frac{1}{min}[/tex] ([tex]k \approx 1.20\,\frac{\%}{min}[/tex]), g) [tex]k \approx 2.876\times 10^{-5}\,\frac{1}{yr}[/tex] ([tex]k \approx 2.876\times 10^{-3}\,\frac{\%}{yr}[/tex])

Step-by-step explanation:

All radioactive isotopes decay exponentially, the mass of the isotope as a function of time ([tex]m(t)[/tex]), measured in grams, is defined by:

[tex]m(t) = m_{o}\cdot e^{-k\cdot t}[/tex] (1)

Where:

[tex]m_{o}[/tex] - Initial mass of the isotope, measured in grams.

[tex]k[/tex] - Decay rate, measured in [tex]\frac{1}{min}[/tex] or [tex]\frac{1}{day}[/tex] or [tex]\frac{1}{yr}[/tex].

[tex]t[/tex] - Time, measured in minutes, days or years.

We define the decay rate in terms of half-life by using the following expression:

[tex]k = \frac{\ln 2}{t_{1/2}}[/tex] (2)

Where [tex]t_{1/2}[/tex] is the half-life of the isotope, measured in minutes or years.

Now we proceed to determine the missing values:

a) Polonium-200 ([tex]t_{1/2} = 11.5\,min[/tex])

[tex]k = \frac{\ln 2}{11.5\,min}[/tex]

[tex]k \approx 0.0602\,\frac{1}{min}[/tex]

b) Lead-194 ([tex]t_{1/2} = 10.7\,min[/tex])

[tex]k = \frac{\ln 2}{10.7\,min}[/tex]

[tex]k \approx 0.0648\,\frac{1}{min}[/tex]

c) Iodine-125 ([tex]k = 0.0115\,\frac{1}{day}[/tex])

[tex]t_{1/2} = \frac{\ln 2}{k}[/tex]

[tex]t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }[/tex]

[tex]t_{1/2} \approx 60.274\,days[/tex]

d) Kryption-75 ([tex]k = 0.161\,\frac{1}{min}[/tex])

[tex]t_{1/2} = \frac{\ln 2}{k}[/tex]

[tex]t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }[/tex]

[tex]t_{1/2}\approx 4.305\,min[/tex]

e) Strontium-79 ([tex]t_{1/2} = 2.3\,min[/tex])

[tex]k = \frac{\ln 2}{2.3\,min}[/tex]

[tex]k \approx 0.3\,\frac{1}{min}[/tex]

f) Uranium-229 ([tex]t_{1/2} = 58\,min[/tex])

[tex]k = \frac{\ln 2}{58\,min}[/tex]

[tex]k \approx 0.0120\,\frac{1}{min}[/tex]

g) Plutonium-239 ([tex]t_{1/2} = 24100\,yr[/tex])

[tex]k = \frac{\ln 2}{24100\,yr}[/tex]

[tex]k \approx 2.876\times 10^{-5}\,\frac{1}{yr}[/tex]

Please help w/ all and leave a small explanation please! :)

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