(a) (i) A radioactive sample produces 1536 decays per second at one point in an
ID: 1401403 • Letter: #
Question
(a) (i) A radioactive sample produces 1536 decays per second at one point in an experiment and 55.888 y later produces 320 decays per second. What is the half- life of this sample? (ii) What s the decay constant (in y^-1)? (iii) What is the mean lifetime of the material? (b) What fraction of a sample of phosphorus-32 is left after 64.26 d. (the half-life of phosphorus-32 is 14.28 d) (e) What period of time is required for a sample of radioactive sulfur-35 to lose 59.75% of its activity? (The half-life of this sample s 87.4 days)Explanation / Answer
a) N = No e^(-yt)
320 = 1536 e^ (-y * 55.888 )
ln(320 / 1536 ) = - y * 55.888
y = 0.0280
for half life
0.5No = No e^( - 0.0280 t)
t = 24.76 years
ii) decay constant = 0.0280 y^-1
iii) mean life time = y / ln2 = 0.0404 y-1
b) half life = 0.693 / decay constant
decay constant = 0.693 / 14.28 = 0.0485 day-1
fraction = N /No = e^(-yt) = e^(-0.0485 x 64.26) = 0.0443
c) half life = 0.693 / decay constant
decay constant = 0.693 / 87.4 = 0.00793day-1
fraction = N /No = e^(-yt)
0.5975 = e^(-0.00793x t)
t = 64.94 days
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