Nuclei of a radioactive element are produced at a constant rate alpha per second. The element has decay constant lambda. Initially N₀ nuclei are present. Find the number of nuclei N at time t.

(1/lambda) * [alpha - (alpha - N₀*lambda)*e^(-lambda*t)]
(1/lambda) * (alpha - N₀*lambda) * e^(-lambda*t)
(alpha - N₀) * e^(-lambda*t)
alpha*N₀*(1 - e^(-lambda*t))

Correct answer: (1/lambda) * [alpha - (alpha - N₀*lambda)*e^(-lambda*t)]

Solution

The ODE dN/dt = alpha - lambda*N has solution N(t) = alpha/lambda + (N₀ - alpha/lambda)*e^(-lambda*t) = (1/lambda)*[alpha - (alpha - N₀*lambda)*e^(-lambda*t)], which matches option A.

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