Half Lives and Radioactive Decay

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SUMMARY

The discussion centers on calculating the number of molybdenum-93 molecules in a meteor and determining how long ago it struck Earth. The calculations yield approximately 3.884 x 1022 molecules currently present in the meteor and indicate that it struck Earth approximately 6079.3 years ago. The decay constant (γ) is derived from the half-life (T1/2) using the formula γ = ln(2)/T1/2, which is essential for solving the decay equations presented.

PREREQUISITES
  • Understanding of radioactive decay and half-life concepts
  • Familiarity with the decay constant and its relationship to half-life
  • Basic knowledge of Avogadro's number and mole calculations
  • Proficiency in logarithmic functions and their applications in decay equations
NEXT STEPS
  • Study the derivation of the decay constant from half-life using γ = ln(2)/T1/2
  • Explore the application of Avogadro's number in calculating the number of molecules
  • Learn about the implications of radioactive decay in geological dating
  • Investigate other isotopes and their decay processes for comparative analysis
USEFUL FOR

Students studying nuclear chemistry, physics, or geology, particularly those focusing on radioactive decay and its applications in dating meteorites and geological samples.

CharlieW95
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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:
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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:

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