Question

Eight grams of Cu$^{66}$ undergoes radioactive decay and after 15 minutes only 1 g remains. The half-life, in minutes, is then}

• A. $\frac{15\ln 2}{\ln 8}$
• B. $\frac{15\ln 8}{\ln 2}$
• C. $15/8$
• D. $8/15$
• E. $15\ln 2$

Hint:

Use $\frac{N}{N_0}$ directly to avoid extra steps.

Answer 1

The Correct Option is A

Concept:
\[ N = N_0 e^{-\lambda t} \]

Step 1: Given.
\[ \frac{N}{N_0} = \frac{1}{8} \]

Step 2: Apply formula.
\[ \frac{1}{8} = e^{-\lambda t} \Rightarrow \ln(1/8) = -\lambda t \] \[ \lambda = \frac{\ln 8}{15} \]

Step 3: Half-life.
\[ T_{1/2} = \frac{\ln 2}{\lambda} = \frac{15\ln 2}{\ln 8} \]