Question

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

Answer 1

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)`