N = N0/2 = N0e-lambda T1/2 N = N0e-lambda t T1/2 = 0.693/lambda The \"half-life\

Question

N = N0/2 = N0e-lambda T1/2 N = N0e-lambda t T1/2 = 0.693/lambda The "half-life" is inversely proportional to the decay constant. It is easy to show that after a period of, say. 3 half-lives, the number of undecayed nuclei has decreased to N = (1/2)(1/2)(1/2) N0 = 1/8N0. The '"activity" of a large collection of radioactive nuclei is the number of decays per second and is just the time derivative of Eq. (14), dN/dt. Neglecting the minus sign because N is decreasing, the "activity" is activity = lambda N0 e-lambda t = lambda N (decays/second) Starting from equation 15, proceeding through equation 16, show that equation 17 is correct by taking the log of the exponential

Explanation / Answer

N = No e –t

When N = No / 2 then the time taken is known as half life T½

Taking log ( to the base e ) on both sides we get

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.