11. Radioactivity and Series Radioactive Decays

[@mitocw]

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[@shikshokio1]

There is again a mistake in the solution of the differential equation at the beginning of the lecture. (Although the final answer is correct). You cannot absorb the exp(-lambda*t) in the integration constant C, because the former is time dependent, and not a constant!

[@dirklutz2818]

so that's why lambda2 appeared again in the final formula!

[@shikshokio1]

@Out Of Order you know, even MIT professors are only human. 😉

[@Priestonwheel]

Agree...

[@koenth2359]

It is correct, he divided everything by exp (lambda t) as a simplification step *after* the constant of integration was introduced. See also my post with the whole derivation

[@achrafBadiry]

awesome teacher

[@koenth2359]

I found the derivation somewhat hard to read and follow, so I worked it out. I thought it might be helpful to others as well. Here it is: We're solving the d.e.-system relating quantities and activities for parent N₁, radiactive daughter N₂ and its stable daughter N₃. Note his notation: N₁₀ means N₁ at t=0 and is a constant. (1) N₁' = -λ₁N₁ (2) N₂' = λ₁N₁-λ₂N₂ (3) N₃' = λ₂N₂ Solve (1) as N₁= N₁₀exp(-λ₁t) and substitute that solution into (2). Rearrange to get N₂'+λ₂N₂ = λ₁N₁₀exp(-λ₁t) This type of integral can be solved introducing a factor μ(t) = exp(∫..dt), where the integrand is the factor that goes with N₂ in the above, so μ(t) = exp(∫λ₂dt) = exp(λ₂t) Multiply everything by μ N₂'exp(λ₂t) + λ₂N₂exp(λ₂t) = λ₁N₁₀exp((λ₂-λ₁)t) Compact the LHS using the product rule: d N₂exp(λ₂t)/dt = λ₁N₁₀exp((λ₂-λ₁)t) Intregrate both sides N₂exp(λ₂t) = λ₁N₁₀/(λ₂-λ₁) exp((λ₂-λ₁)t) + C Set the constant such that N₂₀ = 0, the exp function evaluates to 1 at t=0, so C= -λ₁N₁₀/(λ₂-λ₁) Fill in that value for C and divide everything by μ again N₂ = λ₁N₁₀/(λ₂-λ₁) (exp(-λ₁t) - exp(-λ₂t)) I checked his solution by plugging it into (2) and it satisfies that. And of course N₃ = N₁₀ - N₁ - N₂, so N₃ = N₁₀ (1 - exp(-λ₁t) - λ₁/(λ₂-λ₁) (exp(-λ₁t) - exp(-λ₂t))) = N₁₀ λ₁λ₂/(λ₂-λ₁) ((1 - exp(-λ₁t)) / λ₁ - (1 - exp(-λ₂t)) / λ₂)

[@ngtszyu]

There is a typo for the subtitle at 18:33, where H2 should be N2 :)

[@mitocw]

Thanks for your note! The caption has been updated.

[@ngtszyu]

@@mitocw you are welcome 😁

[@davidwilkie9551]

When we learn to read an analog Clockface, we are staring at the instantaneous holographic periodicity reintegration of QM-TIME sync-duration connectivity. Radioactive decay is s a property similar to the distribution of Clocking half-life positioning around the planet, some clocks, let's say the Electron-photon-phonon-Proton oscillation in Neutronic zero-infinity sync-duration mass-energy-momentum continuous relative-timing ratio-rates of resonant probability change, are fairly easily observed and can be related to the Planck Dimensions phenomena of AM-FM In-form-ation coherence-cohesion fractalisation in Perspective that is the .dt zero-infinity instantaneous self-definition of this Eternity-now Interval Conception. Congratulations of Primes and Cofactors of time-timing Reciproction-recirculation bubble-modes in Totality are considerably more comprehensible than a universe beginning and ending in nothing, while never exhibiting the indeterminacy of disorder that would imply. This theme is ready for Sciencing Analysis and Development in reviewed, revised and reiterated Curricula.

[@qubex]

Isn’t the physical interpretation of the integration constant for N2 the initial concentration of it if there is some present in the population already?

[@dobotube]

Any guesses as to how far he's going to keep this madness going without saying the word "eigenvalue" ?

[@jacobsimpson6775]

when 0 means 0 is that just rounding on different levels?

[@brainstormingsharing1309]

👍👍👍👍👍

[@dogwithwigwamz.7320]

The Pro hopes his students don`t have fissile material at home. I live in England where we are suspect if we have paracetamol at home.

[@pleyergamer6057]

Poao