How Large is the Activity of I_131 After 2 Days?

  • Thread starter Thread starter soopo
  • Start date Start date
  • Tags Tags
    Activity
Click For Summary
SUMMARY

The activity of I-131 after two days, starting from an initial activity of 0.74 MBq and with a half-life of 8 days, is calculated using the formula A = A_0 e^{-\lambda t}. The decay constant λ is derived from λ = ln(2) / T_0.5. After performing the calculations, the resulting activity is 6.22E5 Bq, but the correct value is 250 kBq, which accounts for the fact that only 40% of the activity in a human is I-131.

PREREQUISITES
  • Understanding of radioactive decay and half-life concepts
  • Familiarity with the decay constant calculation
  • Proficiency in using exponential functions in calculations
  • Knowledge of unit conversions to SI units
NEXT STEPS
  • Study the principles of radioactive decay in detail
  • Learn about the applications of I-131 in medical treatments
  • Explore advanced calculations involving decay chains and multiple isotopes
  • Investigate the impact of biological factors on radioactive decay measurements
USEFUL FOR

Students in nuclear physics, medical physicists, and professionals involved in radiology or nuclear medicine who require a solid understanding of radioactive decay principles and calculations.

soopo
Messages
222
Reaction score
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



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.
 
Physics news on Phys.org
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"
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
6K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K