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What is the nuclear binding energy in joules for an atom of lithium? Assume the following:
Mass defect [tex]6.9986235 \times 10^{-29} \text{ kilograms}[/tex].
Use [tex]E = m c^2[/tex], with [tex]c = 3 \times 10^8 \text{ m/s}[/tex].

A. [tex]6.3 \times 10^{-12}[/tex] joules
B. [tex]0.6352 \times 10^{-8}[/tex] joules
C. [tex]6.3 \times 10^{-8}[/tex] joules
D. [tex]21.124 \times 10^{-21}[/tex] joules


Sagot :

Let's solve the problem step-by-step using the formula for nuclear binding energy [tex]\(E = mc^2\)[/tex].

### Step 1: Write Down the Given Values
- Mass defect ([tex]\(m\)[/tex]): [tex]\(6.9986235 \times 10^{-29}\)[/tex] kg
- Speed of light ([tex]\(c\)[/tex]): [tex]\(3 \times 10^8\)[/tex] m/s

### Step 2: Write Down the Formula
The formula to calculate the nuclear binding energy is:
[tex]\[ E = mc^2 \][/tex]

### Step 3: Substitute the Given Values into the Formula
[tex]\[ E = (6.9986235 \times 10^{-29} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]

### Step 4: Calculate the Energy
First, we need to square the speed of light:
[tex]\[ (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]

Next, multiply this result by the mass defect:
[tex]\[ E = 6.9986235 \times 10^{-29} \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \][/tex]

[tex]\[ E = 6.29876115 \times 10^{-12} \, \text{joules} \][/tex]

### Step 5: Match the Answer With the Choices
Comparing this result with the options provided:

A. [tex]\(6.3 \times 10^{-12}\)[/tex] joules

B. [tex]\(0.6352 \times 10^{-8}\)[/tex] joules

C. [tex]\(6.3 \times 10^{-8}\)[/tex] joules

D. [tex]\(21.124 \times 10^{-21}\)[/tex] joules

The correct answer is:
[tex]\[ 6.29876115 \times 10^{-12} \, \text{joules} \approx 6.3 \times 10^{-12} \, \text{joules} \][/tex]

Thus, the correct answer is:

A. [tex]\(6.3 \times 10^{-12}\)[/tex] joules.