Calculating Half-Life: 0.48g to 0.003g

  • Thread starter Thread starter Zoheb Imran
  • Start date Start date
  • Tags Tags
    Half-life
AI Thread Summary
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
Messages
10
Reaction score
0

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?
 
Physics news on Phys.org
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 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top