Half Lives and Radioactive Decay

Click For Summary

Homework Help Overview

The discussion revolves around the topic of radioactive decay and half-lives, specifically focusing on a problem involving the calculation of the number of molybdenum-93 molecules in a meteor and determining the time since the meteor struck Earth based on its original and current mass.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the decay constant and half-life, with some attempting to derive the time since the meteor struck using the decay formula. Others question the validity of the calculated number of molecules, suggesting that the original poster's answer seems low.

Discussion Status

Some participants have provided insights into the calculations and formulas involved, while others express confusion about specific aspects of the problem. There is an ongoing examination of the assumptions made regarding the number of molecules, with no clear consensus reached yet.

Contextual Notes

Participants note potential issues with the interpretation of the answers provided, particularly regarding the representation of scientific notation and the implications of the calculated values.

CharlieW95
Messages
3
Reaction score
0

Homework Statement



2. A meteor contains 6kg of molybdenum-93. Calculate:
a) The number of molybdenum molecules currently present in the meteor.
b) How long ago the meteor struck the Earth if the original amount contained 20kg of molybdenum.

Homework Equations


Answers:
2. a) 3.884x1022 molecules b) 6079.3 yrs

The Attempt at a Solution


Only topic i don't understand, exam tomorrow and i have no idea where else to go !

Any help is appreciated :)

(The answers were given, i need help working how to solve!)
 
Last edited:
Physics news on Phys.org
Formulas:
N = No e-γt
where the decay constant, γ(gamma), and T1/2 are related by
γ\:T1/2 = ln(2)
 
Well...all you need to do is solve for t.
3.
Now N=0.1*No
0.1*No = No*e-γt
0.1=e-γt
ln 0.1=-yt
t=(ln 0.1)/-y=(ln 0.1)/-(ln2/T_1/2)=(ln 0.1)/-(ln2/3.76x10^5 a)=1249044.964 a≈1.25 million years
 
Thank you both so much, understand 1 and 3 easily now. Question 2 is giving me grief!
 
Are you sure about a? 4 thousand molecules sounds very low.
 
sjb-2812 said:
Are you sure about a? 4 thousand molecules sounds very low.
I am sorry, I didn't quite catch that.
 
CharlieW95 said:

Homework Statement


...Answers:
2. a) 3.884x1022 molecules

The answer to a) has been given as above, which is about 4000 molecules.
 
sjb-2812 said:
The answer to a) has been given as above, which is about 4000 molecules.

copy pasting generally doesn't paste superscripts properly, so I would assume it's 1022
 
sjb-2812 said:
The answer to a) has been given as above, which is about 4000 molecules.

lep11 said:
I am sorry, I didn't quite catch that.

SHISHKABOB said:
copy pasting generally doesn't paste superscripts properly, so I would assume it's 1022

Maybe, but even taking that into account, still seems low.
 
  • #10
3.884*1022 is low?
 
  • #11
Yes. How many molecules are there in a mole
 
  • #12
Not sure if this is correct, but I got N=n*N_A=(m/M)*N_A=(6.0*10^3g/92.90681261g/mol)*6.0221327*10^23 1/mol=3.88914...*10^25
,where N_A=Avogadro constant.
If you understood this, you can solve b easily, right?
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K