Strength of Cobalt-60 Source Due in 7 hours, 31 minutes and research application

Question

Strength of Cobalt-60 Source Due in 7 hours, 31 minutes and research applications. One ofthese is soCo. which has a half-life of 520 years and decays by the emission of a beta photons (energies 1.17 Mev and 1.33 Mev). A sclentist wishes to prepare a 0Co sealed r the activity is 14.7 O, how many Co atoms are in the source? partide (energy 0.310 Mev) and two gamma source that will have an activity of at least 14.7 Cl after 35.0 months of use. What is the minimum number of nuclei in the source at the time of creation? Tries 0/10 What is the minimum initial mass of 0Co required? An Tries 0/10

Explanation / Answer

part 1,

You need to find the number of atoms when the activity is 14.7Ci, so you have to use the equation,
R = L* N ------------------(1) ; L - decay constant,
Since you are not given the value of L, you have to find it using,
L = ln(2) / Half Life -------------------(2)
(1) and (2) gives,
R = ln(2) * N / half_life
=>
N = R * half_life / ln(2)

now substitute values. Here R is given in Ci, so you have to convert it to Bq by multiplying it by 3.7*10^(10), because => 1 Ci = 3.7*10^(10)Bq

N = (14.7 * 3.7*10^(10)) * (5.2*365*24*60*60) / (ln(2))
N = 1.287 * 10^20

Part 2

Now use the equation,
N = No * exp (-Lt)

N is the number we just found, t is the time in seconds (35 months or 35/12 years)

No = N * exp (+Lt)
No = N * exp ( ln(2) * t / half_life )
No = 1.287 * 10^20 * exp (ln (2) * (35/12)/ 5.2)
No = 1.9 * 10^20

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