How do you linearise the radioactive decay function properly?

So I need to put:#N(t) = N_oe^(-t/tau)# into the #y=mx+c# form with identities and I think the #-t/tau# is throwing me out here as i'm used to #lambda# with #k# as a constant.

1 Answer
1s2s2p
Mar 7, 2018

The symbol #tau# is used for the mean lifetime which is equal to #1/lambda#, so #e^(-t/tau)=e^(-t/(1/lambda))=e^(-lambdat)#

#N=N_0e^-(t/tau)#

#ln(N)=ln(N_0e^-(t/tau))=ln(N_0)+ln(e^-(t/tau))#
#color(white)(ln(N))=ln(N_0)-t/tau#

Since #N_0# is a y-intercept, #ln(N_0)# will give a y-intercept.and since #-1/tau# is a constant, and #t# is a variable.

#ln(N)=y#
#ln(N_0)=c#
#t=x#
#-1/tau=m#

#y=mx+c#
#ln(N)=-t/tau+ln(N_0)#

Related questions
Impact of this question
1565 views around the world
You can reuse this answer
Creative Commons License