Question

A.Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 10.6 percent of its original value?
______yrs

B.What is the half-life of nitrogen-18 if 1.19 seconds are required for the activity of a sample of nitrogen-18 to fall to 27.1 percent of its original value?

______sec

Answer 1

time required fr 10.6 drop

half life equation:

A = A0*(1/2)^(t/HL)

A = actual concentration in time "t", t time, A0 initial concentation, HL = half life

so:

A = 10.6/100 of A0 --> 0.106A0

so

0.106*A0 = A0*(1/2)^(t/30.2)

0.106= (1/2)^(t/30.2)

ln(0.106) / ln(0.5) = (t/30.2)

t = 30.2 * ln(0.106) / ln(0.5) = 97.78348

B)

apply same logic as before:

A = A0*(1/2)^(t/HL)

27.1 = 100 * (1/2) * ( 1.19/HL)

ln(0.271) / ln(0.5) = 1.19/HL

HL = 1.19 *  ln(0.5) /ln(0.271)

HL = 0.631 seconds