Question

a 60.0 - g sample of fermium - 253 was placed in a sealed vessel 9.0 days ago. only 7.5 g of this isotope is now left. what is the half - life of fermium - 253?
3.0 days
4.5 days
17.5 days
20.0 days

Explanation:

Step1: Set up decay formula

The radioactive - decay formula is $N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}$, where $N$ is the final amount, $N_0$ is the initial amount, $t$ is the time elapsed, and $T_{1/2}$ is the half - life. Here, $N_0=60.0$ g, $N = 7.5$ g, and $t = 9.0$ days. Substitute these values into the formula: $7.5=60.0(\frac{1}{2})^{\frac{9.0}{T_{1/2}}}$.

Step2: Simplify the equation

First, divide both sides of the equation by 60.0: $\frac{7.5}{60.0}=(\frac{1}{2})^{\frac{9.0}{T_{1/2}}}$. Since $\frac{7.5}{60.0}=\frac{1}{8}$, the equation becomes $\frac{1}{8}=(\frac{1}{2})^{\frac{9.0}{T_{1/2}}}$.

Step3: Rewrite $\frac{1}{8}$ in terms of $\frac{1}{2}$

We know that $\frac{1}{8}=(\frac{1}{2})^3$. So, $(\frac{1}{2})^3=(\frac{1}{2})^{\frac{9.0}{T_{1/2}}}$.

Step4: Equate the exponents

Since the bases are the same ($\frac{1}{2}$), we can set the exponents equal to each other: $3=\frac{9.0}{T_{1/2}}$.

Step5: Solve for $T_{1/2}$

Cross - multiply to get $3T_{1/2}=9.0$. Then, divide both sides by 3: $T_{1/2}=\frac{9.0}{3}=3.0$ days.

Answer:

A. 3.0 days