If N(t)=Ce^{- \\lambda t}, where C is some constant, what is {dN(t)}/{dt}? Expre

Question

If N(t)=Ce^{- lambda t}, where C is some constant, what is {dN(t)}/{dt}?
Express your answer in terms of C, lambda, and t.
{dN(t)}/{dt} =-(lambda)*Ce^(-lambda *t)
Correct

Part B
Given that at time t=0 there is a quantity of nuclei present N_0 (in other words, N(0)=N_0), find the value of the constant C in the equation N(t)=Ce^{- lambda t}.
Express your answer in terms of N_0 and lambda.
C =N_{0}
Correct

Part C
Combining your answers from Parts A and B gives the equation N(t)=N_0e^{- lambda t}. Find the time T_1/2 at which half of the original number of nuclei will remain.
Express your answer in terms of N_0 and lambda.
T_1/2 =

Explanation / Answer

we have,

N = C e^( - t)

at half life,

N = C/2

C/2 = C e^(-t)

1/2 = e^(-t)

taking log ,

-ln2 = -t

half life = t = ln2/ = 0.693/

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