Substance Decay Proportionality Constant

The question describes a substance that decays at a rate proportional to the amount of the substance present at any given time. This type of decay is often referred to as exponential decay, and it is governed by a first-order differential equation.

Let $N(t)$ be the amount of substance present at time $t$. The rate of decay is $\frac{dN}{dt}$. According to the problem, this rate is proportional to the amount $N$. Since it's decay, the rate is negative, so we can write the relationship as:

$\frac{dN}{dt} = -kN$

Here, $k$ is the proportionality constant, also known as the decay constant. We need to find the value of $k$.

To solve this differential equation, we can separate variables:

$\frac{dN}{N} = -k \, dt$

Integrating both sides:

$\int \frac{dN}{N} = \int -k \, dt$

$\ln|N| = -kt + C$

Exponentiating both sides (assuming $N > 0$):

$N(t) = e^{-kt + C} = e^C e^{-kt}$

Let $N_0 = e^C$, which represents the initial amount of substance at time $t=0$. So, the solution is:

$N(t) = N_0 e^{-kt}$

Substance Half-Life and Decay Constant

The problem states that half of the substance decays in one year. This is the definition of the half-life ($T_{1/2}$) of the substance. The half-life is the time required for the substance to decay to half of its initial amount.

Given $T_{1/2} = 1$ year.

At $t = T_{1/2}$, the amount of substance remaining is $N(T_{1/2}) = \frac{N_0}{2}$.

Substitute these values into the solution equation:

$\frac{N_0}{2} = N_0 e^{-kT_{1/2}}$

Divide both sides by $N_0$ (assuming $N_0 > 0$):

$\frac{1}{2} = e^{-kT_{1/2}}$

Calculating the Proportionality Constant

We have the equation $\frac{1}{2} = e^{-kT_{1/2}}$. To find $k$, we take the natural logarithm (logarithm base $e$) of both sides:

$\ln\left(\frac{1}{2}\right) = \ln(e^{-kT_{1/2}})$

Using logarithm properties, $\ln(a^b) = b \ln(a)$ and $\ln(e) = 1$, and $\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$:

$\ln(1) - \ln(2) = -kT_{1/2} \ln(e)$

$0 - \ln(2) = -kT_{1/2} \cdot 1$

$-\ln(2) = -kT_{1/2}$

Multiply both sides by $-1$:

$\ln(2) = kT_{1/2}$

Now, we are given that the half-life $T_{1/2}$ is 1 year. Substitute $T_{1/2} = 1$ into the equation:

$\ln(2) = k \cdot 1$

$k = \ln(2)$

The natural logarithm $\ln(2)$ is the same as $\log_e 2$.

Thus, the proportionality constant is $\log_e 2$. This matches the first option.

Options Review

• Option 1: $\log_e 2$. This matches our calculated value for the proportionality constant $k$.
• Option 2: $\log_{10} 2$. This is the base-10 logarithm, which is different from the natural logarithm.
• Option 3: $(\log_e 2)/2$. This value would result if the half-life was 2 years instead of 1.
• Option 4: $(log 10 2 )/2$. This option uses the base-10 logarithm and an incorrect half-life value (implied 2 years).

The correct proportionality constant is $\log_e 2$ when the half-life is 1 year.