The amount of radioactive substance after 38 days

Click For Summary
SUMMARY

The discussion focuses on calculating the remaining amount of a radioactive substance after 38 days, given a half-life of 3.8 days and an initial mass of 10.24 grams. The decay constant, k, is derived using the formula k = ln(2)/3.8, resulting in k = 0.1824. The remaining amount after 38 days is calculated using the exponential decay formula A = 10.24 * e^(-0.1824 * 38), yielding approximately 0.01 grams. The discussion also clarifies the notation used for multiplication and decimal points in mathematical expressions.

PREREQUISITES
  • Understanding of exponential decay and half-life concepts
  • Familiarity with natural logarithms and their properties
  • Knowledge of the mathematical constant e (approximately 2.718)
  • Ability to interpret mathematical notation and symbols
NEXT STEPS
  • Study the derivation of the exponential decay formula in detail
  • Learn about radioactive decay and its applications in various fields
  • Explore the use of natural logarithms in solving real-world problems
  • Investigate different methods for calculating decay constants
USEFUL FOR

Students in physics or chemistry, educators teaching radioactive decay, and professionals in fields involving nuclear science or environmental studies.

Pushoam
Messages
961
Reaction score
53

Homework Statement


upload_2017-12-12_14-30-30.png


Homework Equations

The Attempt at a Solution



Is there anyway to choose the correct option without solving it?

## \lambda = \frac { 0.693 } { t_{\frac 1 2}} ##
[/B]
## N = N_0 e^{- 0.693 {\frac { t } { t_{\frac 1 2 }}}} ##

## N = 10.24 e^{- \frac { 0.693 * 38 } { 3.8} } = 10^{-2} ~ gm##So, the correct option is (b).

Is this correct?
 

Attachments

  • upload_2017-12-12_14-30-30.png
    upload_2017-12-12_14-30-30.png
    3.4 KB · Views: 839
Physics news on Phys.org
Why the raised dot between those digits? If to mean decimal point, that is not the way to show it.
Half-life, 3.8 days;
Want how much after 38 days;
Begin with 10.24 grams

If model is for exponential decay, then A=10.24*e^(-k*t).
Find k using half-life information.
(1/2)=1*e^(-k*3.8)
(1/2)=e^(-3.8k)
ln(1/2)=ln(e^(-3.8k))
ln(1/2)=-3.8k*1
k=-ln(1/2)/3.8
k=ln(2)/3.8
k=0.1824

Revise for the more specific decay equation:
A=10.24*e^(-0.1824t)

How much remaining after 38 days?
Let t=38 and evaluate A.
 
38 days is 10 half-lives. 2^10 = 1024. 10.24/1024 = 0.01.
 
  • Like
Likes   Reactions: Pushoam
symbolipoint said:
Why the raised dot between those digits? If to mean decimal point, that is not the way to show it.
It is the symbol of multiplication. Please, see the keyboard for this. You, too, have used it in the following expression,
symbolipoint said:
A=10.24*e^(-k*t).

symbolipoint said:
k=ln(2)/3.8
I have written ##\lambda## instead of k.
The above is nothing but, k = 0.693 /t½.
 
Pushoam said:
It is the symbol of multiplication. Please, see the keyboard for this. You, too, have used it in the following expression,
I have written ##\lambda## instead of k.
The above is nothing but, k = 0.693 /t½.
I looked at the linked document you posted in the beginning post. There was a raised dot but that is to be used to signify multiplication. We want the character to be at the botton of the digits levels to read it as a decimal point. If I use a raised dot, I use it as a multiplication operation. That is the normal, common usage. The middle quoted passage shows the asterisk also often used for multiplication, and common and usually understood when using plain text to write an algebraic expression; it might appear as a raised dot, or it might appear exactly the asterisk. The 10.24 shows the decimal point, and the "three point eight" as the decimal number should be written as 3.8 (notice the dot at the bottom-level).
 
  • Like
Likes   Reactions: Pushoam

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K