your bow tie helped me listen better btw. i also love that you showed the few times you made a mistake and how you found out it was a mistake, and humbly corrected it. seeing someone do something completely perfect 100% of the time doesn't help people failing at the subject 90% of the time. so just seeing you forget the negative sign or even finishing the equation on the wrong line helped me see if i were to make a mistake, HOW to correct it! your teaching is more valuable than you may know and i wanted you to know that.
Sir at 1:17 why did you multiply negative lambda with the right side of your equation?How, is it some kind rule? I am from cie a levels they never taught us anything like ddt you did on both sides. I think i have asked a very dumb question.
Not a dumb question at all. Sometimes we just don't see something. Happens all the time to me as well. When taking the derivative of an exponential function: d/dt (e^-at) = -a e^-at
The minus simply indicates that the number of radioactive atoms is decreasing. But when we want to express the number of decays per second, it doesn't make any sense to keep the negative sign. (you can't have negative decays per second)
The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ. (Copied off of Britannica)
If I at the beginning get the decay rate of something, let's say 3070 decays/s in 500g of a sample, and carbon provides 225bq/kg, when you have to calculate the age of the sample, do you always have to take the ratio of the sample to the carbon one?
If the radioactive decay rate is the slope of the N(t) curve why is the value computed to be a constant since the slope changes (becomes less steep) as t increases?
Reply to myself. Yes the rate changes with time but the video is computing the CURRENT rate which is based on the sample size. This works fine for Carbon (half life 5730 years) but not radon (half life in days). Another question: For RANDOM carbon sample it would still have to be a "sutiable" sample - meaning for example organic or not carbon close to a source that would make the C14/C12 ratio not 1/10^12 (e.g. deep underground sample or sample affected by other radioactive sources, etc.)
Yo, The decay RATE is a variable and varies as a function of time. N(t). The decay CONSTANT (lambda) is indeed a constant and determines the steepness of the exponential curve. y = e^(2X) Lambda is like the number 2 in the equation. The decays constant is NOT the decay rate.
I swear this guy knows everything
your bow tie helped me listen better btw. i also love that you showed the few times you made a mistake and how you found out it was a mistake, and humbly corrected it. seeing someone do something completely perfect 100% of the time doesn't help people failing at the subject 90% of the time. so just seeing you forget the negative sign or even finishing the equation on the wrong line helped me see if i were to make a mistake, HOW to correct it! your teaching is more valuable than you may know and i wanted you to know that.
Sir at 1:17 why did you multiply negative lambda with the right side of your equation?How, is it some kind rule?
I am from cie a levels they never taught us anything like ddt you did on both sides.
I think i have asked a very dumb question.
Not a dumb question at all. Sometimes we just don't see something. Happens all the time to me as well. When taking the derivative of an exponential function: d/dt (e^-at) = -a e^-at
this was really insightful, thank you sir!
Glad it was helpful!
Excellent!
Glad you liked it!
Sir, where does the minus sign go in the final answer?
The minus simply indicates that the number of radioactive atoms is decreasing. But when we want to express the number of decays per second, it doesn't make any sense to keep the negative sign. (you can't have negative decays per second)
@@MichelvanBiezen another way that the minus sign is used for is on logarithms and finding the reciprocal of a value
You are my Savior ! Thank you very much ! Very clear !
Where did you get the 0.693?
The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T1/2, and the decay constant is given by T1/2 = 0.693/λ. (Copied off of Britannica)
@@thomaswilson1107 correct
ln(2)
If I at the beginning get the decay rate of something, let's say 3070 decays/s in 500g of a sample, and carbon provides 225bq/kg, when you have to calculate the age of the sample, do you always have to take the ratio of the sample to the carbon one?
Yes. You have to use the ratio of the mass of your sample to the mass of the object for which you know the reference decay rate.
Nice, thanks for the answer.
Hey! i thought all your videos on this topic are very clear and straightforward! Also arranged well, and i revised pretty well from them, thanks a lot
You are welcome. Thanks for the comment.
If the radioactive decay rate is the slope of the N(t) curve why is the value computed to be a constant since the slope changes (becomes less steep) as t increases?
Reply to myself. Yes the rate changes with time but the video is computing the CURRENT rate which is based on the sample size. This works fine for Carbon (half life 5730 years) but not radon (half life in days).
Another question: For RANDOM carbon sample it would still have to be a "sutiable" sample - meaning for example organic or not carbon close to a source that would make the C14/C12 ratio not 1/10^12 (e.g. deep underground sample or sample affected by other radioactive sources, etc.)
Yo,
The decay RATE is a variable and varies as a function of time. N(t).
The decay CONSTANT (lambda) is indeed a constant and determines the steepness of the exponential curve.
y = e^(2X) Lambda is like the number 2 in the equation.
The decays constant is NOT the decay rate.
Why did you not mutiply e^-lambda/t to the last equation . That also consists of the rate of decays ?!
We did. N(t) is the function with the exponent.
@@MichelvanBiezen but you just multiplied the no. Of nucleons and not the exponent in your calculations
@@khushijain4969 Yes, when t = 0, the exponential part equals 1.
legend
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is this GCSE or A level
It is not specifically geared towards any specific test, but the videos cover the key concepts of all physics.
How is carbon 14 one trillion of carbon 12
Why is trillion?
it is the relative abundance