-Iridium-192 is one radioisotope used in brachytherapy, in which a radioactive s

Question

-Iridium-192 is one radioisotope used in brachytherapy, in which a radioactive source is placed inside a patient's body to treat cancer. Brachytherapy allows the use of a higher than normal dose to be placed near the tumor while lowering the risk of damage to healthy tissue. Iridium-192 is often used in the head or breast. Answer the following three questions (a, b, and c) based on the radioactive decay curve of iridium-192, shown below. Click on the graph and move the mouse over a point to get values for that point.

a) If the initial sample is 6.25 g, what mass of the original iridium-192 remains after 80 days?

b) Estimate the half-life of the radioisotope

c) How many days would it take for three-fourths of the sample to decay?

Explanation / Answer

I have used the graph from another source but I couldn't get exact values because of lack of facility of moving the mouse over a point to get values for that point. So the values may not be exact , please adjust accordingly.

(a)

N(t) = N(0) * e^( - lambda * t)

No = 6.25 g

After 80 days around 48% of sample is remaining

So,

mass of original iridium-192 remaining = 0.48 * 6.25 g = 3 g

(b)

t(1/2) = ln(2) / lambda

t(1/2) = 0.6931 / lambda

So we will calculate lambda first

N(t) = N(0) e^(- lambda * t )

N(t)/N(0) = e^( - lambda * t)

- lambda * t = ln[ N(t) / N(0) ]

lambda = - 1/t * ln[ N(t) / N(0) ]

lambda = - 1/80 * ln ( 3 / 6.25)

lambda = 9.2 x 10^-3 day^-1

So,

t(1/2) = 0.6931 / 9.2 x 10^-3

t(1/2) = 75.55 days

(c)

3/4th of sample decays means 1/4 th of sample is remaining

So,

N(t) / N(0) = 1/4 = 0.25

From earlier equation

lambda = - 1/t * ln[N(t) / N(0)]

So,

t = - 1 / lambda * ln[ N(t) / N(0) ]

t = - 1 / 9.2 x 10^-3 * ln(0.25)

t = 150.68 days

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