Question

problem 6
a certain radioactive material follows the law of exponential change and has a half - life of 38 hours. find how long it takes for 90% of the radioactivity to be dissipated.
ans. 126 hours
problem 7
irene has php10,200 and she is planning to deposit it on a mysterious bank where the interest is compounding continuously with a rate of 4.2%. how much would be the increase from the initial amount after 5 years?
a. php12,583.52
b. php8,583.52
c. php6,383.52
d. php2,383.52

Explanation:

Problem 6

Step1: Write the radioactive - decay formula

The formula for radioactive decay is $N = N_0e^{kt}$, where $N$ is the amount of the material at time $t$, $N_0$ is the initial amount, $k$ is the decay constant, and $t$ is the time. When $t = T_{1/2}$ (half - life), $N=\frac{N_0}{2}$. So, $\frac{N_0}{2}=N_0e^{kT_{1/2}}$. Canceling out $N_0$ gives $\frac{1}{2}=e^{kT_{1/2}}$. Taking the natural logarithm of both sides: $\ln(\frac{1}{2}) = kT_{1/2}$, and $k=\frac{\ln(\frac{1}{2})}{T_{1/2}}$. Given $T_{1/2}=38$ hours, $k = \frac{\ln(0.5)}{38}$.

Step2: Find the time when 90% is dissipated

If 90% of the radioactivity is dissipated, then the remaining amount $N=(1 - 0.9)N_0 = 0.1N_0$. Substitute into the decay formula $N = N_0e^{kt}$: $0.1N_0=N_0e^{kt}$. Cancel out $N_0$ to get $0.1 = e^{kt}$. Take the natural logarithm of both sides: $\ln(0.1)=kt$. Substitute $k=\frac{\ln(0.5)}{38}$ into $\ln(0.1)=kt$: $t=\frac{\ln(0.1)}{\frac{\ln(0.5)}{38}}=\frac{38\ln(0.1)}{\ln(0.5)}$.
Calculate $\frac{38\ln(0.1)}{\ln(0.5)}=\frac{38\times(- 2.3026)}{-0.6931}\approx126$ hours.

Step1: Use the continuous - compounding formula

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. Given $P = 10200$, $r=0.042$, and $t = 5$.
So, $A=10200e^{0.042\times5}=10200e^{0.21}$.

Step2: Calculate the final amount

$e^{0.21}\approx1.2337$. Then $A = 10200\times1.2337=12583.54$.

Step3: Calculate the increase

The increase is $A - P$. So, $12583.54−10200 = 2383.54\approx2383.52$.

Answer:

126 hours

Problem 7