Question

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Answer 1

Answer:

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

Step-by-step explanation:

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

$m(t) = m_{o}\cdot e^{-k\cdot t}$ (1)

Where:

$m_{o}$ - Initial mass of the isotope, measured in grams.

$k$ - Decay rate, measured in $\frac{1}{min}$ or $\frac{1}{day}$ or $\frac{1}{yr}$.

$t$ - Time, measured in minutes, days or years.

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

$k = \frac{\ln 2}{t_{1/2}}$ (2)

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

Now we proceed to determine the missing values:

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

$k = \frac{\ln 2}{11.5\,min}$

$k \approx 0.0602\,\frac{1}{min}$

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

$k = \frac{\ln 2}{10.7\,min}$

$k \approx 0.0648\,\frac{1}{min}$

c) Iodine-125 ($k = 0.0115\,\frac{1}{day}$)

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

$t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }$

$t_{1/2} \approx 60.274\,days$

d) Kryption-75 ($k = 0.161\,\frac{1}{min}$)

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

$t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }$

$t_{1/2}\approx 4.305\,min$

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

$k = \frac{\ln 2}{2.3\,min}$

$k \approx 0.3\,\frac{1}{min}$

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

$k = \frac{\ln 2}{58\,min}$

$k \approx 0.0120\,\frac{1}{min}$

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

$k = \frac{\ln 2}{24100\,yr}$

$k \approx 2.876\times 10^{-5}\,\frac{1}{yr}$