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Ivan
3 years ago
6

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
Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

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.

DanielleElmas [232]3 years ago
3 0

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_1is 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.

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