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Need answer thanks! only answer if you know please​!-example-1

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Answer:

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

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
(1)/(min) or
(1)/(day) or
(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 = (\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 = (\ln 2)/(11.5\,min)


k \approx 0.0602\,(1)/(min)

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


k = (\ln 2)/(10.7\,min)


k \approx 0.0648\,(1)/(min)

c) Iodine-125 (
k = 0.0115\,(1)/(day))


t_(1/2) = (\ln 2)/(k)


t_(1/2) = (\ln 2)/(0.0115\,(1)/(day) )


t_(1/2) \approx 60.274\,days

d) Kryption-75 (
k = 0.161\,(1)/(min))


t_(1/2) = (\ln 2)/(k)


t_(1/2) = (\ln 2)/(0.161\,(1)/(min) )


t_(1/2)\approx 4.305\,min

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


k = (\ln 2)/(2.3\,min)


k \approx 0.3\,(1)/(min)

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


k = (\ln 2)/(58\,min)


k \approx 0.0120\,(1)/(min)

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


k = (\ln 2)/(24100\,yr)


k \approx 2.876* 10^(-5)\,(1)/(yr)

User Jszpilewski
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