2- A radioactive substance decays at a rate proportional to itself. There are 200 grams at the start and 150 grams at the end of 12 hours. A) Find the amount A

Question

2- A radioactive substance decays at a rate proportional to itself. There are 200 grams at the start and 150 grams at the end of 12 hours.
A) Find the amount A present at any time t. [Round k to four decimal places]
B) How much will remain after 30 days? [Round your answer to three significant digits.]
C) How many days will it take for the number of grams to 25 grams? [Round your answer to three significant digits.]

Answer 1

General Equation for radioactive decay:

 

N=N₀e^kt

 

A) We seek to find a value for k

 

e^kt = (N/N₀)

 

We are given N = 150 g, N₀ = 200 g and t = 12 hours or 0.5 days

 

e^kt = 0.75

 

Take ln of each side

 

kt = ln(0.75)

 

k = -0.240 (if we express t in terms of hours)

 

or k = -0.5754 (if we express t in terms of days) (12  hours = 0.5 days

 

B) N=N₀e^kt

 

N = (200 g)e^{(-0.5754)(30 days)}

 

e^{(-0.5754)(30 days)} = N/200

 

Take ln of both sides

 

(-0.5754)(30 days) = ln(N) - ln(200)

 

ln(N) = (-0.5754)(30 days) + ln(200)

 

ln(N) = -11.9626

 

N = 6.378 x 10^-6 g

 

C) N=N₀e^kt

 

N = 25 g, evaluate t

 

e^kt = (25 g)/(200 g)

 

ln(e^kt) = kt = ln(0.125)

 

(-0.5754)t = -2.0794

 

t = (-2.0794)/(-0.5754)

 

t = 3.61 days

 

Abdu, please check my arithmetic.  I think it's right, but who knows.

 

 

 

 

Answer 2

A) Decay law: N = 200 e^-kt

Find k from 150 = 200 e^-k(12) ⇒ k = -(1/12) ln (150/200) ≈ 2.397*10^-2/hr

N = 200 (0.75)^t/12hr ≈ 200 e^{-0.02397 t/hr}

 

B) At t=30 d = 60*12 hr = 720 hr, N = 200 (0.75)⁶⁰ ≈ 6.38*10^-6 g

 

C) 25 = 200 (0.75)^t/12 ⇒  t = ln(25/200) /(-k) = 12 ln (25/200) / ln(0.75) ≈ 86.7 hr = 3.61 days

Answer 3

Let A be the amount present in grams at time t in days and let r be a constant.

 

Decay at a rate proportional to itself can be modeled as A'=rA

 

A'/A=r

 

Integrating both sides with respect to t:

 

ln(A) = rt + c

 

where c is a constant. Let C = exp(c) and exponentiate both sides.

 

A(t)=C*exp(rt)

 

A(0) = C =  200

 

A(t)=200*exp(rt)

 

A(0.5) = 200*exp(0.5r) = 150

exp(0.5r)=150/200

0.5r=ln(0.75)

r=2*ln(0.75)≈-0.5754

 

A) A(t) = 200*exp(-0.5754t)

 

B) A(30) = 200*exp(-0.5754*30) ≈6.38 * 10^-6 grams

 

C) 25 = 200*exp(-0.5754t)

     exp(-0.5754t) = 0.125

     -0.5754t = ln(0.125)

     t = ln(0.125)/(-0.5754) ≈ 3.61 days

 

 

Answer 4

A) The equation is as follows:

 

dm/dt=-km, where k is a decay constant. The solution of this equation is:

 

m(t)=m(0)*e^-kt

 

Thus, if we had m(0)=200 and m(12)=150, then 150=200*e^-12k, from which k=1/12*ln(200/150)=1/12*ln(4/3)≈0.0240 [ hour^-1]. So, the amount present at any time is:

 

m(t)=200*e^-0.024t, where m is in grams, t is in hours.

 

B) 30 days equals to 30*24=720 hours. The amount left will be:

 

m(720)=200*e^-0.024*720≈6.38*10^-6 gram.

 

C) If the amount left is 25 grams, then 25=200*e^-0.024*t, from which

t=1/0.024*ln(200/25)=1/0.024*ln(8)=1/0.008*ln(2)≈86.643 hours or t≈3.61 days.

Answer 5

A)

   A = A₀e^kt

 

 

 

150 = 200e^12k

 

ln(3/ 4) = e^12k

 

k = ln(0.75)/12 = -0.024

 

  A = 200 e ^-0.024t

 

 B)   A = 200e^-0024(720) = 6.26* 10^-6= 6.26 μgr

 

C)   25 = 200 e ^-0.024t

 

      0.125 = e^0.024t

 

 

   t = ln(0.125)/ -0.024 = 86.6 hrs

 

      t = 86.6/24 = 3.61 days.

 

Answer 6

Can I assume you have some calculus background?

"A radiactive substance decays at a rate proportional to itself."

This means

 

rate = kQ

where Q is the amount of the substance at a given time.

In calculus terms,

 

dQ/dt = -kQ

 

Note that it is negative because the rate of change should be negative. The amount of stuff is decreasing.

The solution for these decaying problems is an exponential function.

Q(t) = Q₀e^-kt

(If you are in precalculus, then the solution should have been given to you; it's just left to you to figure out what k is.)

(A) You can use the condition that, given a starting amount of 200 g (Q₀=200) there is 150g after 12 hours (Q(0.5) = 150, where I've chosen to express time in days) to solve for k:

150 = 200*e^-k(0.5)

 

0.75 = e-k(0.5)

ln 0.75 = -k(0.5)

k = -(ln 0.75)/0.5 ~ 0.5754

(B) Now that you have a formula for the quantity remaining after any given time:

Q(t) = Q₀e^-0.5754t

you can solve this problem.

Q(30) = (200 g)e^-0.5754(30)

 

Q(30)~6.38*10^-6 g.

 

(C) In this case, you have the final quantity, and need to solve for time.

25 g = (200 g)e^-0.5754t

 

0.125 = e^-0.5754t

ln 0.125 = -0.5754t

t = -(ln 0.125)/0.5754 ~ 3.61 days