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The starting materials in a nuclear change have a total mass of [tex]2.3465 \times 10^{-27} \, \text{kg}[/tex]. After the change, the resulting materials have a mass of [tex]2.3148 \times 10^{-27} \, \text{kg}[/tex]. How much energy was released during the change?

A. [tex]2.85 \times 10^{-12} \, \text{J}[/tex]

B. [tex]2.08 \times 10^{-10} \, \text{J}[/tex]

C. [tex]2.11 \times 10^{-10} \, \text{J}[/tex]

D. [tex]8.56 \times 10^{-4} \, \text{J}[/tex]


Sagot :

To determine how much energy was released during the nuclear change, we need to use the famous equation from Einstein's theory of relativity, which is [tex]\( E = mc^2 \)[/tex]. Here, [tex]\( E \)[/tex] represents the energy released, [tex]\( m \)[/tex] stands for the mass difference between the initial and final states, and [tex]\( c \)[/tex] is the speed of light (approximately [tex]\( 3 \times 10^8 \)[/tex] meters per second).

1. Calculate the mass difference:
- Initial mass: [tex]\( 2.3465 \times 10^{-27} \)[/tex] kg
- Final mass: [tex]\( 2.3148 \times 10^{-27} \)[/tex] kg

[tex]\[ \text{Mass difference} = 2.3465 \times 10^{-27} \, \text{kg} - 2.3148 \times 10^{-27} \, \text{kg} \][/tex]

[tex]\[ \text{Mass difference} = 3.17 \times 10^{-29} \, \text{kg} \][/tex]

Here, we have rounded the mass difference for simplicity.

2. Calculate the energy released using the mass difference and the speed of light [tex]\( c \)[/tex]:

[tex]\[ E = (3.17 \times 10^{-29} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2 \][/tex]

Evaluate the speed of light squared:

[tex]\[ (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, (\text{m}^2/\text{s}^2) \][/tex]

Now multiply this by the mass difference:

[tex]\[ E = (3.17 \times 10^{-29} \, \text{kg}) \times (9 \times 10^{16} \, (\text{m}^2/\text{s}^2)) \][/tex]

Thus, the energy released [tex]\( E \)[/tex]:

[tex]\[ E = 2.853 \times 10^{-12} \, \text{J} \][/tex]

Comparing this result to the provided options:

1. [tex]\( 2.85 \times 10^{-12} \, \text{J} \)[/tex]
2. [tex]\( 2.08 \times 10^{-10} \, \text{J} \)[/tex]
3. [tex]\( 2.11 \times 10^{-10} \, \text{J} \)[/tex]
4. [tex]\( 8.56 \times 10^{-4} \, \text{J} \)[/tex]

The energy released during the nuclear change is [tex]\( 2.85 \times 10^{-12} \, \text{J} \)[/tex], which matches the first answer provided.

Therefore, the correct answer is:
[tex]\[ \boxed{2.85 \times 10^{-12} \, \text{J}} \][/tex]