Calculating 'half-life' *help : /

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SUMMARY

The discussion focuses on calculating the half-life of a radioactive isotope based on its decay rates over a specified time period. The initial decay rate is 8305 decays per minute, decreasing to 3070 decays per minute over 5 days. The half-life (T1/2) is determined using the formula T1/2 = ln(2)/lambda, where lambda is calculated from the decay rates using the equation ln(R1/R2) = lambda * delta t. The final calculated half-life is approximately 3.48 days.

PREREQUISITES
  • Understanding of radioactive decay concepts
  • Familiarity with the natural logarithm function
  • Knowledge of the decay constant (lambda)
  • Basic algebra for solving equations
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  • Learn about the significance of the decay constant (lambda) in nuclear physics
  • Explore practical applications of half-life calculations in various fields
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This discussion is beneficial for students in physics or chemistry, educators teaching radioactive decay, and anyone interested in understanding half-life calculations in nuclear science.

MrChaos
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Hello :)

I came across a question and I haven't been able to solve it so far. It would be great if maybe someone could give me a few tips.

Here is the question:

Measurements of the radioactivity of a certain isotope tell you that the decay rate
decreases from 8305 decays per minute to 3070 decays per minute over a period of 5.00
days.

What is the half-life T1/2 of this isotope?

How can I get the right answer for this question? I never really worked with those half-life
questions. I am still interested how it works.

thanks :)
 
Last edited:
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Hi MrChaos and welcome to PF. Please observe our rules and use the template for posting requests for help with homework.

What equation do you know that relates activity to time and half-life>
 
umm...

N = N0 e-lambda*t

i'm not sure if this ones right

T1/2= ln 2/lambda
 
mission impossible now complete
on my last attempt i got it right

ln (R_1/R_2)= lambda*deltat
ln (8305/3070)= lambda*5days
0.995180181= lambda*5days
0.995180181*1/5 = 0.19903636 = lambda

lambda*T= ln 2= 0.6931
0.6931/0.19903636= T
T_1/2= 3.482283978
= 3.48 days

:)
 

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