Recent content by Sir Bro

Ah right. I still did it right though didn't I?

N=No e^-λt N/2=No e^-λt 208.5= 417* e^-0.0357t 0.5= e^-0.0357t ln(0.5)= -0.0357t t= 19.416! YES! Thank you very much haruspex and ehild :) I thought when taking the natural log you had to do it to both sides.

N=No e^-λt N/2=No e^-λt 208.5= 417* e^-0.0357t 0.5= e^-0.0357t 0.5= ln(-0.0357t) 0.5= t * -ln(0.0357) 0.15=t?

Thanks for the reply. I've learned something from this at least but I think I'll leave it out of my report; I believe this is above my knowledge and I don't want to dive into something and waste even more time than I already have on this assignment lol. Thanks again mate :)

Oh yeah whoops, just a typo. How do I get the natural log of a negative number? My calculator returns "nonreal answer"

Aha. Would you believe me if I said I already had. Contacted the World Nuclear association regarding a different question about a week ago but didn't speak to anyone "genius". The guy was more than helpful but his qualification was purely a bachelors of science so not much help there for...

Here's what I have now: N=No e^-λt N/2=No e^-λt 208.5= 417* e^0.0357t 0.5= e^0.0357t ln(0.5)= ln(0.0357t) ln(0.5)= t * ln(0.0357) t= 0.208? :/

So full working would be: Just wait; No/2 = No e-λt 2/2= 2 * e-λt 2 = 4 * e-λt (did u just miss the negative lambda exponent or am i missing something?)"ln2=λt" shouldn't it be this? ln(2) = ln(-λt)What is the equation I need to start with? Maybe from there I could get a better idea of...

Understood. That gives a correct answer which is good but... Not only do I want to understand what is going on here for my own benefit but I need to be able to understand what is happening here so I can reproduce it in the form of working. Sorry for my ignorance regarding logarithms; we...

Can you explain that in a bit more detail please? I don't really understand why I'm supposed to use the reciprocal of the gradient. t = [1/ -0.0357] * ln(2) ? That also gives a wrong answer I don't understand :/

So use -0.0375 as the decay constant in becquerels (represented by lambda). This ends up giving a result of 8.21 minutes. That's a very long way off the actual half life?

So I had a look into this and found this equation: Where is the mass-averaged absorbed dose of the entire item T is the item of interest is the absorbed dose as a function of location is the density as a function of location is volume Is it even possible for me to calculate the absorbed...

Homework Statement Decay of Carbon-11 from given data of count rate at time 0 to count rate at time 14 minutes in 2 minute intervals. Homework Equations I feel like I've gone wrong here because the answer I get of 19.6 minutes is actually negative if I go back and do the last equation again. I...

I'm writing a report on the feasibility of nuclear energy and one of the things I wanted to address is the issue of radiation exposure risk from waste when stored correctly and incorrectly; in water pools compared to say just left above ground. Thanks for the reply. I'll look into gray's in a...

Homework Statement So I'm looking at the radiation exposure risk regarding depleted uranium (U-238) and nuclear waste (4.20% U-235, 95.8% U-238). Yes I know these compositions are incorrect; they're close enough to exact for my purposes. The only information I have is the half lives of the...