half life question?

Question

a sample of tritim-3 decayed to 94.5% of its original amount after a year
a) What is the half-life of tritium-3?
b) How long would it take the sample to decay to 20% of its original amount?

Answer 1

Hmm, ok Lets bring up the equation T1/2=(ln(1/y))/(ln(2)) where t1/2 is the number of half lives. ln is natural log. Y is the percentage of stuff left.
I got
t1/2= (ln(1/0.945))/0.693
which means T1/2 = 0.0816 that is how many half lives are in one year. Now lets see how many years make up one half life.

1/0.0816=12.255 years per half life. So that's the half life.

For question B lets look at another equation.

This time it will be

Tage=((halflife)/(0.693))*ln(1/y)
Tage=((12.255)/(0.693))*ln(1/0.20)
Tage=28.454 years to get the sample to 20%

Answer 2

a)The reaction is a first order reaction, so you can use the equation:
ln(c0/c)=kt
c0 is the initial concentration
c is the concentration at a certain moment
k is the rate constant or rate coefficient, a value dependent on temperature
t is the period of time in which the concentration increased from c0 to c

You must replace c0 with 100, and c with 94.95 in the relationship: ln(100/94.5)=k*1. In this way, you can determine the constant k=0.05657, and the unit of measure for k is 1/year.
Half-life is the time it takes for the concentration to fall from c0 to c0/2, so you have to replace, in the first relationship, c with c0/2. You obtain: ln2=k*t. From this relationship you can determine the half-life: t=(ln2)/k=(ln2)/0.05657=12.25 years.
b)You can find the period of time in which the sample decays to 20% of its original amount in a similar way: you have to replace c0 with 100 and c with 20:
ln(100/20)=k*t => t=(ln5)/k=(ln5)/0.05657=28.45 years.

Hope this helped. Good luck and remember that kinetics is easy, you just have to memorise some relationships and apply them.

Answer 3

Why don't you figure it out, it's ; e^-lam*t