Math Is Fun Forum

fusilli_jerry89

Member

Registered: 2006-06-23

Posts: 86

Quick question about Deriving radioactive decay equations

Just a quick question. When you have the rate of change of radioactive decay as                         dN/dt=-(lambda)N.

Then can someone please explain to me why we rearrange this equation to: dN/N=-(lambda)dt.
I don't get how this integrates into N=(No)e^(-(lambda)t). Partially because I havn't done math for the summer, and because I don't get why we put change in N divided by N. Wouldn't change in N divided by N equal the fraction of decrease at any given time? Does this have anything to do with lambda?

also, can some1 plz explain how integrating dv/(g-bv/m) = dt gives u (-m/b)ln((mg-bv)/mg) = t?

Thank you

Last edited by fusilli_jerry89 (2007-09-19 14:57:57)

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Quick question about Deriving radioactive decay equations

Try this one:

Given that

---

X'(y-Xβ)=0

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Quick question about Deriving radioactive decay equations

lambda refers to the wavelength of a photon, usually in [Å], and E refers to its energy, usually in [keV].

Actually, lambda is the symbol used to represent the radioactiv decay constant, and in this case does not mean wavelength.