Suppose a material has half-life 30 minutes, and suppose initially we have 10g. 1. How much remains after 20 minutes? 2. How long will it take for 2g to remain?

Question

Please show me how you got your answers.

Answer 1

Dale,

 

Let t      = elapsed time in minutes since the start

Let T     = the established half-life for this material in minutes = 30 min.

Let m(t) = amount of material at time t, where m(0) = 10g initially

 

You can model m(t) using the equation:     m(t) = m(0)[1/2]^(t/T). 

Note that when t=T=30min, (t/T) =1 and m(T) = m(0)[1/2]¹ = m(0)/2 so half is gone at the half-life time.

Substituting in the data:

m(t) = 10[1/2]^(t/30).

 

So after 20 min, t/T = 20/30 = 0.667, and m(20) = 10[0.5]^(0.667) = 10[0.6298) = 6.298g left

 

For 2g to remain, find t such that m(t) = 2 = 10[0.5]^(t/30).

Taking logarithms to base 10 on both sides of = sign,  log2 = log(10) + (t/30)log (0.5),

or 30 log2 = 30 log10 + t log(0.5)

30(0.3010) = 30(1) + (-0.3010) t

9.03 = 30 - 0.3010 t , so t = [30.0 - 9.03]/0.3010 = 20.97/0.3010 = 69.68 min or about 69.98/T = 2.32 half lives.

 

The amounts left after each succeeding 30 minutes = 5g 2.5g 1.25g so it falls between 2 & 3 and checks!

 

Answer 2

If you plan on going into engineering, it is good practice to consider this an exponentially decaying problem like radioactive decay. It is good practice to model this as

 

N(t)=N_o*e^kt 

 

Where N_o is the initial amount of substance at zero time

k is the constant of proportionality related to the rate in which the substance decays.

t = the time lapse

 

Note: The reason why this is a good model for this problem is that it is a solution to the differential equation as being the time rate of change of a substance is proportional to the given amount of substance.

 

The first step in working out this problem is to solve for the value k when the half-life of a substance is 30 minutes.

 

1/2*N_o=N_o*e^k*30 Which implies 1/2=e^30*k 

 

Take the natural lag of both sides

 

ln(1/2)=ln(e^30*k) This implies -ln(2)=30*k

 

Thus, k= -ln(2)/30 = -0.0231/min

 

Hence,

 

N(t)=10*e^-0.0231*t

 

For t=20 minutes

 

N(20)=10*e^-0.0231*20 = 6.30 grams

 

To figure out how long it will take for the substance to decay to 2 grams is:

 

2=10*e^-0.0231*t The way I make this problem easier is I divide everything by 10 in which becomes

 

2/10=1/5=e^-0.0231*t

 

Now take the natural log of both sides in which becomes

 

ln(1/5)=-0.0231*t   This implies t=69.67 minutes.

Answer 3

N = N₀/ 2^η

N is material left after η half lives

 

N₀ is original material

 

half lives = time/half life = 30/20 = 1.5

 

N = 10/2^1.5 = 3.535g

 

Similarly to find time

 

2 = 3/2^η

 

2^η = 3/2

 

n log2 = log 3 - log 2

 

n = number of half lives=  (0.47712 - 0.30103)/(0.30103) = 0.58496

 

time elapsed = η x number of half lives = 17.5488 minutes