Calculating 'half-life' *help : /

[MrChaos]

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 :)

 

[kuruman]

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>

 

[MrsChaos]

umm...

N = N₀ e^-lambda*t

i'm not sure if this ones right

T_1/2= ln 2/lambda

 

[MrsChaos]

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

:)

 

[rdl]

Hello!

Calculating the half-life of an isotope involves using the decay rate and the time period over which the decay occurs. In this case, we have a starting decay rate of 8305 decays per minute and an ending decay rate of 3070 decays per minute over a period of 5.00 days. To calculate the half-life, we can use the following equation: T1/2 = (ln 2)/λ.

First, we need to find the decay constant (λ) by using the decay rate and the time period. We can do this by using the following equation: λ = (ln N2/N1)/t, where N2 is the final decay rate, N1 is the initial decay rate, and t is the time period. Plugging in our values, we get: λ = (ln 3070/8305)/5.00 = -0.0862 days^-1.

Now, we can plug this value into the equation for half-life to get: T1/2 = (ln 2)/(-0.0862 days^-1) = 8.05 days. This means that the half-life of this isotope is 8.05 days.

I hope this helps and provides some guidance on how to approach half-life calculations. Let me know if you have any further questions. Good luck!