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Activity of I_131 after 2 days

  • Thread starter soopo
  • Start date
  • #1
225
0

Homework Statement


The initial activity of I_131 is 0.74MBq.
The half time of I_131 is 8 days.

How large is the activity after two days?


Homework Equations



[tex] A = A_0 e^{-\lambda t} [/tex]


The Attempt at a Solution



We know
t = 2 days
A_0 = 0.74 MBq
T_0.5 = 8 days

1. Solve the activity constant
[tex] \lambda = ln2 / T_0.5 [/tex]

2. Plug it to the equation
[tex] A = A_0 e^{(-ln2 / T_0.5) * t} [/tex]

I standardise the units to SI and then omit/cancel them
[tex] A = 0.74E6 * e^{-ln2 / 4} [/tex]
[tex] = 6.22E5 Bq [/tex]

---

The right answer is 0.4 times what I get
[tex] A = 0.4 * 6.22E5 Bq [/tex]
[tex] = 250 kBq[/tex]

I am not sure where the 0.4 is got.
 

Answers and Replies

  • #2
225
0

Homework Statement


The initial activity of I_131 is 0.74MBq.
The half time of I_131 is 8 days.

How large is the activity after two days?


Homework Equations



[tex] A = A_0 e^{-\lambda t} [/tex]


The Attempt at a Solution



We know
t = 2 days
A_0 = 0.74 MBq
T_0.5 = 8 days

1. Solve the activity constant
[tex] \lambda = ln2 / T_0.5 [/tex]

2. Plug it to the equation
[tex] A = A_0 e^{(-ln2 / T_0.5) * t} [/tex]

I standardise the units to SI and then omit/cancel them
[tex] A = 0.74E6 * e^{-ln2 / 4} [/tex]
[tex] = 6.22E5 Bq [/tex]

---

The right answer is 0.4 times what I get
[tex] A = 0.4 * 6.22E5 Bq [/tex]
[tex] = 250 kBq[/tex]

I am not sure where the 0.4 is got.
The problem is now solved.

There is the following sentence two pages before the exercise
"Only 40% of the activity in a human is of the form I_131"
 

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