Question

The​ half-life of the radioactive element unobtanium dash 37 is 10 seconds. If 32 grams of unobtanium dash 37 are initially​ present, how many grams are present after 10 ​seconds? 20 ​seconds? 30 ​seconds? 40 ​seconds? 50 ​seconds?
Question content area bottom
Part 1
The amount left after 10 seconds is
  
enter your response here grams.

Answer 1

To calculate the amount of a radioactive substance left after a certain time, we can use the formula:

\[
A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}
\]

Where:
- \( A \) is the amount remaining after time \( t \)
- \( A_0 \) is the initial amount
- \( t_{1/2} \) is the half-life
- \( t \) is the total time elapsed

In this case:
- \( A_0 = 32 \) grams
- \( t_{1/2} = 10 \) seconds

We can calculate the remaining amount after each specified time.

**1. After 10 seconds:**
\[
A = 32 \left( \frac{1}{2} \right)^{\frac{10}{10}}
\]
\[
A = 32 \left( \frac{1}{2} \right)^1
\]
\[
A = 32 \times 0.5 = 16 \text{ grams}
\]

**2. After 20 seconds:**
\[
A = 32 \left( \frac{1}{2} \right)^{\frac{20}{10}}
\]
\[
A = 32 \left( \frac{1}{2} \right)^2
\]
\[
A = 32 \times 0.25 = 8 \text{ grams}
\]

**3. After 30 seconds:**
\[
A = 32 \left( \frac{1}{2} \right)^{\frac{30}{10}}
\]
\[
A = 32 \left( \frac{1}{2} \right)^3
\]
\[
A = 32 \times 0.125 = 4 \text{ grams}
\]

**4. After 40 seconds:**
\[
A = 32 \left( \frac{1}{2} \right)^{\frac{40}{10}}
\]
\[
A = 32 \left( \frac{1}{2} \right)^4
\]
\[
A = 32 \times 0.0625 = 2 \text{ grams}
\]

**5. After 50 seconds:**
\[
A = 32 \left( \frac{1}{2} \right)^{\frac{50}{10}}
\]
\[
A = 32 \left( \frac{1}{2} \right)^5
\]
\[
A = 32 \times 0.03125 = 1 \text{ gram}
\]

So the final answers are:
- After 10 seconds: **16 grams**
- After 20 seconds: **8 grams**
- After 30 seconds: **4 grams**
- After 40 seconds: **2 grams**
- After 50 seconds: **1 gram**