Question

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How much of the sample remains after 120 years? (Round your answer to mg (c) After how long will only 1 mg remain? (Round your answer to one decimal place.) T = yr

Answer 1

(a) The mass remains after t years is
`180.42 e^t` mg.

(b)Therefore the remains sample after 120 years is 11.25 mg.

(c)Therefore after 224.76 years only 1 mg will remain.

Explanation:

The differential equation of decay

`(dN)/(dt)=-kN`

`\Rightarrow (dN)/(N)=-kdt`

Integrating both sides

`\int (dN)/(N)=\int-kdt`

`\Rightarrow ln|N|=-kt+c_1`

`\Rightarrow N=e^(-kt+c_1)` [
`c_1`is arbitrary constant ]

`\Rightarrow N=ce^(-kt)`
`[ e^(c_1)=c ]`

Initial condition is,
`N=N_0` when t=0

`\Rightarrow N_0=ce^(-k.0)`

`\Rightarrow N_0= c`

Therefore
`N=N_0e^(-kt)`........(1)

N= Amount of radioactive material after t unit time.

`N_0`= initial amount of radioactive material

k= decay constant.

Half life:

`N= \frac12N_0` , t= 30 years

`\therefore \frac12 N_0= N_0e^(-k* 30)`

`\Rightarrow \frac12=e^(-30t)`

`\Rightarrow -30k= ln|\frac12|`

`\Rightarrow k= (ln|\frac12|)/(-30)`

`\Rightarrow k=(ln|2|)/(30)`

(a)

The mass remains after t years N.

`N_0=180`

Now we put the value of
`N_0` in the equation (1)

`\therefore N=180 * e^-(ln`

`\Rightarrow N= 180* e^(-0.023t)`.........(2)

The mass remains of cesium after t years is
`180* e^(-0.023t)` mg.

(b)

Putting
`N_0=180` and t=120 years in equation (2)

`N=180e^(-0.023*120)`

`\Rightarrow N=11.25`

Therefore the remains sample after 120 years is 11.25 mg.

(c)

Now putting N= 1 in equation (2)

`1= 180* e^(-0.023t)`

`\Rightarrow (1)/(180)=e^(-0.023t)`

Taking ln both sides

`\Rightarrow ln|(1)/(180)|=ln|e^(-0.023t)|`

`\Rightarrow -ln|180|=-0.023t`

⇒t=224.76 (approx)

Therefore after 224.76 years only 1 mg will remain.

Answer 2

a)
`Q(t) = 180e^(-0.023t)`

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Explanation:

The following equation is used to calculate the amount of cesium-137:

`Q(t) = Q(0)e^(-rt)`

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

`Q(t) = Q(0)e^(-rt)`

`0.5Q(0) = Q(0)e^(-30r)`

`e^(-30r) = 0.5`

Applying ln to both sides of the equality.

`\ln{e^(-30r)} = ln(0.5)`

`-30r = ln(0.5)`

`r = (ln(0.5))/(-30)`

`r = 0.023`

So

`Q(t) = Q(0)e^(-0.023t)`

180-mg sample, so Q(0) = 180

`Q(t) = 180e^(-0.023t)`

(b) How much of the sample remains after 120 years?

This is Q(120).

`Q(t) = 180e^(-0.023t)`

`Q(120) = 180e^(-0.023*120)`

`Q(120) = 11.4`

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

`Q(t) = 180e^(-0.023t)`

`1 = 180e^(-0.023t)`

`e^(-0.023t) = (1)/(180)`

`e^(-0.023t) = 0.00556`

Applying ln to both sides

`\ln{e^(-0.023t)} = ln(0.00556)`

`-0.023t = ln(0.00556)`

`t = (ln(0.00556))/(-0.023)`

`t = 225.8`

225.8 years.