Calculating Half-Life: 0.48g to 0.003g

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To determine the amount of a radioactive substance present 20 hours before testing, one must calculate backwards using the half-life of 5 hours. Since 20 hours corresponds to four half-lives, the calculation involves doubling the remaining amount for each half-life. Starting from 0.48g, the correct progression is 0.48g to 0.96g after one half-life, and continuing this process results in an initial amount of 0.96g before the sample was tested. The confusion arose from calculating the decay instead of the original amount prior to testing. Understanding the direction of time in calculations is crucial for accurate results.
Zoheb Imran
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Homework Statement



The Half-life of a radioactive substance is 5 hours. A sample is tested and found to contain 0.48g of the substance.
How much of the substance was present in the sample 20 hours before the sample was tested?

Homework Equations





The Attempt at a Solution


i did like this...

20/5 = 4 half life

0.48---->0.24--->0.12--->0.06--->0.003g amount of substance was present.

am i right?
 
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Zoheb Imran said:
How much of the substance was present in the sample 20 hours before the sample was tested?

Right idea, but you want the other direction in time...
 
why the other direction? can you please explain? that's where i am having problem...in calculating...
 
Zoheb Imran said:
why the other direction? can you please explain? that's where i am having problem...in calculating...

You are calculating forward in time when you used the half-life. So you are figuring out how much of the radioactive material is left after it was sampled. The question asks you to figure out how much of the material was present before the sample was made...
 
thanks!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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