Question

find the half-life (in hours) of a radioactive substance that is reduced by 5 percent in 10 hours.
half life = (include units)

Explanation:

Step1: Define decay formula

The radioactive decay formula is $N(t) = N_0 e^{-kt}$, where $N(t)$ is remaining quantity, $N_0$ is initial quantity, $k$ is decay constant, $t$ is time.

Step2: Substitute given values

After 10 hours, 95% remains: $0.95N_0 = N_0 e^{-k(10)}$. Cancel $N_0$: $0.95 = e^{-10k}$.

Step3: Solve for decay constant $k$

Take natural log of both sides: $\ln(0.95) = -10k$. Rearrange: $k = -\frac{\ln(0.95)}{10} \approx \frac{0.051293}{10} = 0.0051293$.

Step4: Use half-life condition

For half-life $t_{1/2}$, $0.5N_0 = N_0 e^{-kt_{1/2}}$. Cancel $N_0$: $0.5 = e^{-kt_{1/2}}$.

Step5: Solve for half-life

Take natural log: $\ln(0.5) = -kt_{1/2}$. Rearrange: $t_{1/2} = -\frac{\ln(0.5)}{k}$. Substitute $k$:
$t_{1/2} = -\frac{\ln(0.5)}{-\frac{\ln(0.95)}{10}} = 10 \cdot \frac{\ln(0.5)}{\ln(0.95)}$
Calculate: $t_{1/2} \approx 10 \cdot \frac{-0.6931}{-0.051293} \approx 10 \cdot 13.51 = 135.1$

Answer:

135.1 hours