How Large is the Activity of I_131 After 2 Days?

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The initial activity of I_131 is 0.74 MBq with a half-life of 8 days. After two days, the calculated activity using the decay formula is approximately 622,000 Bq. However, the expected correct answer is 250 kBq, which is 40% of the calculated value. This discrepancy arises from the fact that only 40% of the activity in a human is attributed to I_131. The problem has been resolved by clarifying this percentage.
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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



A = A_0 e^{-\lambda t}

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
\lambda = ln2 / T_0.5

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

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

---

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

I am not sure where the 0.4 is got.
 
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soopo said:

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



A = A_0 e^{-\lambda t}


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
\lambda = ln2 / T_0.5

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

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

---

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

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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