Radioactive Decay: Nuclear Radiation: Edexcel A-level Physics
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- Опубликовано: 25 авг 2024
- Specification points 172 and 173 (141, 142 for IAL) detailing radioactivity over time. Including use of exponential equations and their ln equivalents.
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This is AMAZING - exactly how my School teaches it.
I tend to prefer watching videos when I learn as I can better understand it then.
However, in School - I tend to miss out on minor details - so watching videos like these are extremely helpful.#
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Hi, could you explain why my answer is different from yours for the past paper question with the 1600 year half life. I changed the 1600 years into s’¹ and substituted it to get 9.49*10^10 nuclei. May I know if me changing the time into seconds is wrong. The t half I got was 1.37*10^-11 s’¹
There is no way for me to check that like this I’m afraid.
hey how did u get the value of lambda while calculating the question under Geological Dating?
also could you explain the whole calculation part as well
You are given the half-life of uranium so lambda is ln2/half-life
You have to assume that any lead present is the final product of the uranium decay series. So the ratio of lead to uranium tells you how much of the uranium has decayed.
You know that N uranium = N0 uranium x e^-lt (l = lambda)
N0 uranium is current uranium + current lead.
N uranium = (N uranium + N lead) x e^-lt and then we divide across by N uranium.
Leaving 1 = (1 + N lead/Nuranium) x....
N lead/N uranium = 0.7.
So.. 1 = (1 + 0.7) x e....
1/1.7 = e...
ln both sides and you solve for t
Does that help?
@@nutshellphy thank you so much. Really does help. Your channel and effort saved my grades. I switched from Cambridge to Edexcel and honestly I was lost until I found this channel.
Honestly a very underrated channel. Keep up the good work!
Thank you! Much appreciated.
how did u get 5x10*-18 at 16:05
I converted the half-life into seconds and then did ln2 divided by that. Just to have the decay constant in per second. You could leave your half-live in years, get the decay constant in per year, and end up with an answer of t directly in years. I just wanted to show it with seconds. So as long as you keep track of your time unit, it will come out the same way.
@@nutshellphy thank uu ^ ^
sir what is A/Ao???
Ao is the activity when the material/sample started to decay and A is it’s activity after a given time, t. A/Ao is one divided by the other.
@@nutshellphy then why do we divide?
Because it becomes a fraction so you can see what proportion is left and then use that to figure out how old the sample is. Either by deduction (ie. If 1/8 is left, that’s 3 half-lives) or by finding the ln of both sides and then finding a value for t in the equation (assuming you know lambda through the half-life).
@@nutshellphy count rate of 150min^-1 means what
Count rate or activity? If activity, it means the sample is decaying at 150 ‘atoms’ per minute. If count rate, it means your detector is detecting 150 radioactive decay particles entering it per minute.