Half Life Problem: How much time has elapsed?

Question

The half-life of a radioactive substance is 24 days. You begin with 45 grams and you now have 10 grams remaining. How much time has elapsed?

Please show the work so I can understand the reason why.

Answer 1

The radioactive half-life for a given radioisotope is the time for half the radioactive nuclei in any sample to undergo radioactive decay. After two half-lives, there will be one fourth the original sample, after three half-lives one eight the original sample, and so forth.

The half-life formula is: A = (Ao)e^(-0.693 t/T)

In the given problem

T=24 days

Ao=45 grams

A = 10 grams

You are asked to determine t in the formula.

A = (Ao)e^(-0.693 t/T)

45 = 10 e^(-0.693 t/24)

4.5 =  e^- (0.693 t/24)

Solve for t

ln(4.5) = 0.693 t/24

1.5(24)/0.693 =t

t= 51.95

t=52 days

Answer 2

37.33 Days You lose 22.5 grams in 24 days or .9375g a day

45-10= 35g Lost 35/.9375=37.33days

Answer 3

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Answer 4

I don't have a calculator on me but the formula is

A = A(0)*e^(kt)

A is the amount you have now (10 g)

A(0) is the amout you started with (45 g)

e is a constant roughly rounded to 2.71828183

k is the half-life constant

and t is the variable for time elapsed that you're trying to find

Answer 5

Radioactive decay are of first order kinetics

t=0.693/k

t=24days

k=0.693/24=0.02 /days

R(intial value)=45 g

R"(final value)=10 g

k=(2.303/t) x log([R]/[R"]).

=(2.303/ 0.02) x log(45/10)

=75.2 days

Answer 6

Half Life = Half the amount in one passing of a halflife (in this case 24 days)

45 grams - 0 days

22.5 grams - 24 days

11.25 grams - 48 days

To get to 10, you know when it reaches another 24 days, it would have lost 5.625, so take 5.625/24 days which means it loses .234375 grams per day.

So by just guestimating, you know it takes atleast 5 more days

so you can say that 53 days passed, unless it asks for exact answer

Answer 7

10/45 = x/24

cross multiply.

x=5.3333

Answer 8

Since half the mass decays every half life time period: after 24 days you have 1/2 your original mass, after 48 days(2 half-lifes) you have 1/4 your original mass and so on:

To summarize:

45 -> 22.5 after 24 days

22.5 -> 11.25 after 48 days

So it take over 48 days.

The equation is just the starting mass time a power of 2. you want the power to be multiples of 2 for every multiple of the half life so it will have the form t/24 (if t is a multiple of 24 you will get 1, 2, 3 etc). And the power will be negative since you want multiples of (1/2) and not 2.

So it can be written as:

M = M0*2^(-t/24)

at t = 0 ..... M = M0

at t = 24 ... M = M0*2^(-1) = M0/2

at t = 48 ... M = M0*2^(-2) = M0/4

10 = 45*2(-t/24)

10/45 = 2^(-t/24)

log(10/45) = (-t/24)log(2)

t = (-24)log(10/45)/log(2)

t = 52.08 days

Answer 9

The formula for this type of problem is given by

A = Ao(e^kt)

where

A = amount of substance remaining at any time "t"

Ao = initial amount of substance = 45 g (given)

k = proportionality constant

t = time

The above equation can be rewritten as

A = 45(e^kt)

and at t = 24 days (half life), (A/Ao) = 1/2,

Therefore,

(1/2) = e^24k

To solve for "k", take the natural logarithm of both sides,

ln (1/2) = 24k[ln e] and since ln e = 1,

k = (1/24)*ln (1/2)

k = -0.02888

Hence your decay equation is now modified to

A = 45(e^-0.02888t)

and if A = 10, then

10 = 45(e^-0.02888t)

(10/45) = e^-0.02888t

Again, taking the natural logarithm of both sides,

ln (10/45) = -0.02888t (ln e)

Solving for "t",

t = -(1/0.02888)*ln (10/45)

t = 52 days

Answer 10

Here is the entire derivation and solution.

Let:

dQ/dt = the rate of change

Q = the quantity at any time

Qo = the initial quantity

dQ/dt = -kQ. whre k = constant of proportionality

dQ/Q = -kdt

lnQ = -kt + lnC. Where lnC = constant of integration

lnQ - lnC = -kt

ln(Q/C) = -kt

Q/C = e^(-kt)

Q = Ce^(-kt)

If t = 0, Q = Qo, Hence C = Qo. Thus

ln(Q/Qo) = -kt

If t = 24, Q =Qo/2, hence;

ln(1/2) = -24k

k = 0.693/24

Therefore;

t= -ln(Q/Qo)/(0.693/24) = -ln(10/45)/(0.693/24)

t = 52.0893 years