Question

Determine the time required for an original 24.0-gram sample of sr-94 to decay until only 1.5 grams of the sample remains unchanged.

Answer 1

Answer: Check explanation.

Step-by-step explanation:

In this question, we are to calculate for the time required for an original mass of a sample to to decay to another mass of the sample.

We can alive this by using the equation below;

N(t) = N(o) × (1/2)^t/t-half life------------------------------------------------------------(1).

Where N(t/) is the amount left after time t, which is given as 1.5 gram from the question.

N(o) is the initial amount of the substance, this is given as 24 gram.

t-half life= half life if the decaying substance. We make an assumption that the Half life is 10 years.

t= is the time required for an original mass of a sample to to decay to another mass of the sample.

Therefore, 1.5 g/24 g= (1/2)^t/10 years.

=> 0.625 = (1/2)^t/10.

==> t/10 log 0.5 = log 0.625.

==> t = 6.781 years.

Check attached picture for the exclusive solution.

Answer 2

For the answer to the question above,
24 --> 12 --> 6 --> 3 --> 1.5
.....1 .......2 ......3 ......4 ............. four half-lives

The time is 4(t½) ...... where t½ is the half-life.

There are two types of problems dealing with half-life. One is the "super-simple" which is what you have here and the solution is also simple. It involves counting half-lives.

A more realistic problem wouldn't work out so nicely and would involve this equation or a variation since radioactive decay is a first-order process.

A = Ao e^(-kt) ...... where A is the activity (or amount) after some time (t), Ao is the original activity (or amount) at t=0, k is the decay constant and t is the elapsed time. The decay constant is related to the half-life. t½=ln2/k or k=ln2/(t½).

A = Ao e^(-kt)
ln A = -kt + lnAo
ln(A/Ao) = -kt
t = (ln(A/Ao)) / -k
t = ln(1.5/24) / -k
t = ln(0.0625) / -k
t = -2.773 / -k
t = -2.773 / -ln2 / (t½)
t = -2.773 / (-0.693 / (t½))
t = 4(t½)