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

Answer:

a) $Q(t) = 180e^{-0.023t}$

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

c) 225.8 years.

Step-by-step 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.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

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

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

Answer 2

Answer:

(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.

Step-by-step explanation:

The differential equation of decay

$\frac{dN}{dt}=-kN$

$\Rightarrow \frac{dN}{N}=-kdt$

Integrating both sides

$\int \frac{dN}{N}=\int-kdt$

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

$\Rightarrow N=e^{-kt+c_1}$             [ $c_1$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

$\therefore \frac12 N_0= N_0e^{-k\times 30}$

$\Rightarrow \frac12=e^{-30t}$

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

$\Rightarrow k= \frac{ln|\frac12|}{-30}$

$\Rightarrow k=\frac{ln|2|}{30}$

(a)

The mass remains after t years N.

$N_0=180$

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

$\therefore N=180 \times e^{-\frac{ln|2|}{30}t}$

$\Rightarrow N= 180\times e^{-0.023t}$.........(2)

The mass remains of cesium after t years is $180\times e^{-0.023t}$ mg.

(b)

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

$N=180e^{-0.023\times120}$

$\Rightarrow N=11.25$

Therefore the remains sample after 120 years is 11.25 mg.

(c)

Now putting N= 1 in equation (2)

$1= 180\times e^{-0.023t}$

$\Rightarrow \frac{1}{180}=e^{-0.023t}$

Taking ln both sides

$\Rightarrow ln|\frac{1}{180}|=ln|e^{-0.023t}|$

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

⇒t=224.76 (approx)

Therefore after 224.76 years only 1 mg will remain.