Question

Radium decomposes at the rate proportional to the amount present at any time. If $P\%$ of amount disappears in one year, then amount of radium left after 2 years is

• A. $\left(\frac{10 - P}{10}\right)^2$
• B. $x_0 \left[1 + \frac{P}{100}\right]^2$
• C. $x_0 \left[1 - \frac{P}{100}\right]^2$
• D. $x_0 \left[\frac{10 - P}{10}\right]^2$

Hint:

Radioactive decay acts exactly like compound interest in reverse (depreciation)! For a constant annual decay rate of $r = \frac{P}{100}$, the formula for the remaining mass after $n$ intervals is always $x_n = x_0(1 - r)^n$. Substituting $n = 2$ gives the answer immediately!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem describes radioactive decay where the rate of decomposition follows a first-order linear differential equation. We are given that a certain percentage $P\%$ of the substance disappears in the first year, and we need to determine the expression for the mass remaining at the end of 2 years.

Step 2: Key Formula or Approach:
Let $x$ be the mass of radium present at any time $t$. The differential equation modeling this decay behavior is:
$$\frac{dx}{dt} = -kx \implies x(t) = x_0 e^{-kt}$$ Where $x_0$ is the initial quantity present at $t = 0$. Since decay occurs at a uniform fractional rate each year, we can also model this discrete compounding process over time intervals using percentage reductions.

Step 3: Detailed Explanation:
Let the initial mass of radium at $t = 0$ be $x_0$.
1. After 1 year ($t = 1$), $P\%$ of the initial amount decays away. The amount that disappears is $\frac{P}{100} \times x_0$. Therefore, the remaining amount of radium left after 1 year ($x_1$) is:
$$x_1 = x_0 - \frac{P}{100}x_0 = x_0\left(1 - \frac{P}{100}\right)$$ 2. During the second year ($t = 2$), the same constant fraction $P\%$ decomposes, but this time it applies to the remaining quantity $x_1$ rather than the original starting mass. The amount left after 2 years ($x_2$) is:
$$x_2 = x_1 - \frac{P}{100}x_1 = x_1\left(1 - \frac{P}{100}\right)$$ Substitute the expression for $x_1$ into this equation:
$$x_2 = \left[x_0\left(1 - \frac{P}{100}\right)\right] \times \left(1 - \frac{P}{100}\right)$$ $$x_2 = x_0\left(1 - \frac{P}{100}\right)^2$$ This algebraic derivation corresponds perfectly with option (C).

Step 4: Final Answer:
The amount of radium left after 2 years is $x_0 \left[1 - \frac{P}{100}\right]^2$, which corresponds to option (C).