nuclear energy released from a fission mass defect calculation Given mass defect (dm) = 0.0006 kg; speed of light (c) = 300,000,000 m/s, determine the energy released (E) in J.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except E is given, so isolate E symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \Delta m\,c^2$$

3. Step 3

   $$List the givens: mass defect (dm) = 0.0006 kg, speed of light (c) = 300,000,000 m/s$$

4. Step 4 — Substitute the given values:

   $$E = \Delta m\,300000000^2$$

5. Step 5 — Evaluate:

   $$E = 51840000000000\ \text{J}$$

6. Step 6 — Check: returning E = 51,840,000,000,000 J to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 103,680,000,000,000 — kept a factor of two that cancels in the correct rearrangement.
• 25,920,000,000,000 — dropped that same factor in the other direction.
• 57,024,000,000,000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear power plant energy output from mass-energy equivalence Given speed of light (c) = 300,000,000 m/s; energy released (E) = 34,778,000,000,000 J, determine the mass defect (dm) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except dm is given, so isolate dm symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for dm:

   $$dm = \dfrac{E}{c^2}$$

3. Step 3

   $$List the givens: speed of light (c) = 300,000,000 m/s, energy released (E) = 34,778,000,000,000 J$$

4. Step 4 — Substitute the given values:

   $$dm = \dfrac{34778000000000}{300000000^2}$$

5. Step 5 — Evaluate:

   $$dm = 0.0004\ \text{kg}$$

6. Step 6 — Check: returning dm = 0.0004 kg to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0008 — kept a factor of two that cancels in the correct rearrangement.
• 0.0002 — dropped that same factor in the other direction.
• 0.0004 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear reaction energy release estimated from mass defect Given mass defect (dm) = 0.0002 kg; speed of light (c) = 300,000,000 m/s, determine the energy released (E) in J.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except E is given, so isolate E symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \Delta m\,c^2$$

3. Step 3

   $$List the givens: mass defect (dm) = 0.0002 kg, speed of light (c) = 300,000,000 m/s$$

4. Step 4 — Substitute the given values:

   $$E = \Delta m\,300000000^2$$

5. Step 5 — Evaluate:

   $$E = 15570000000000\ \text{J}$$

6. Step 6 — Check: returning E = 15,570,000,000,000 J to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 31,140,000,000,000 — kept a factor of two that cancels in the correct rearrangement.
• 7,785,000,000,000 — dropped that same factor in the other direction.
• 17,127,000,000,000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear energy released from a fission mass defect calculation Given speed of light (c) = 300,000,000 m/s; energy released (E) = 39,640,000,000,000 J, determine the mass defect (dm) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except dm is given, so isolate dm symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for dm:

   $$dm = \dfrac{E}{c^2}$$

3. Step 3

   $$List the givens: speed of light (c) = 300,000,000 m/s, energy released (E) = 39,640,000,000,000 J$$

4. Step 4 — Substitute the given values:

   $$dm = \dfrac{39640000000000}{300000000^2}$$

5. Step 5 — Evaluate:

   $$dm = 0.0004\ \text{kg}$$

6. Step 6 — Check: returning dm = 0.0004 kg to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0009 — kept a factor of two that cancels in the correct rearrangement.
• 0.0002 — dropped that same factor in the other direction.
• 0.0005 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear power plant energy output from mass-energy equivalence Given mass defect (dm) = 0.0007 kg; speed of light (c) = 300,000,000 m/s, determine the energy released (E) in J.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except E is given, so isolate E symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \Delta m\,c^2$$

3. Step 3

   $$List the givens: mass defect (dm) = 0.0007 kg, speed of light (c) = 300,000,000 m/s$$

4. Step 4 — Substitute the given values:

   $$E = \Delta m\,300000000^2$$

5. Step 5 — Evaluate:

   $$E = 63270000000000\ \text{J}$$

6. Step 6 — Check: returning E = 63,270,000,000,000 J to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 126,540,000,000,000 — kept a factor of two that cancels in the correct rearrangement.
• 31,635,000,000,000 — dropped that same factor in the other direction.
• 69,597,000,000,000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear reaction energy release estimated from mass defect Given speed of light (c) = 300,000,000 m/s; energy released (E) = 62,457,000,000,000 J, determine the mass defect (dm) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except dm is given, so isolate dm symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for dm:

   $$dm = \dfrac{E}{c^2}$$

3. Step 3

   $$List the givens: speed of light (c) = 300,000,000 m/s, energy released (E) = 62,457,000,000,000 J$$

4. Step 4 — Substitute the given values:

   $$dm = \dfrac{62457000000000}{300000000^2}$$

5. Step 5 — Evaluate:

   $$dm = 0.0007\ \text{kg}$$

6. Step 6 — Check: returning dm = 0.0007 kg to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0014 — kept a factor of two that cancels in the correct rearrangement.
• 0.0003 — dropped that same factor in the other direction.
• 0.0008 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear energy released from a fission mass defect calculation Given mass defect (dm) = 0.0010 kg; speed of light (c) = 300,000,000 m/s, determine the energy released (E) in J.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except E is given, so isolate E symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \Delta m\,c^2$$

3. Step 3

   $$List the givens: mass defect (dm) = 0.0010 kg, speed of light (c) = 300,000,000 m/s$$

4. Step 4 — Substitute the given values:

   $$E = \Delta m\,300000000^2$$

5. Step 5 — Evaluate:

   $$E = 89820000000000\ \text{J}$$

6. Step 6 — Check: returning E = 89,820,000,000,000 J to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 179,640,000,000,000 — kept a factor of two that cancels in the correct rearrangement.
• 44,910,000,000,000 — dropped that same factor in the other direction.
• 98,802,000,000,000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear power plant energy output from mass-energy equivalence Given speed of light (c) = 300,000,000 m/s; energy released (E) = 23,801,000,000,000 J, determine the mass defect (dm) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except dm is given, so isolate dm symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for dm:

   $$dm = \dfrac{E}{c^2}$$

3. Step 3

   $$List the givens: speed of light (c) = 300,000,000 m/s, energy released (E) = 23,801,000,000,000 J$$

4. Step 4 — Substitute the given values:

   $$dm = \dfrac{23801000000000}{300000000^2}$$

5. Step 5 — Evaluate:

   $$dm = 0.0003\ \text{kg}$$

6. Step 6 — Check: returning dm = 0.0003 kg to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0005 — kept a factor of two that cancels in the correct rearrangement.
• 0.0001 — dropped that same factor in the other direction.
• 0.0003 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear reaction energy release estimated from mass defect Given mass defect (dm) = 0.0007 kg; speed of light (c) = 300,000,000 m/s, determine the energy released (E) in J.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except E is given, so isolate E symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for E:

   $$E = \Delta m\,c^2$$

3. Step 3

   $$List the givens: mass defect (dm) = 0.0007 kg, speed of light (c) = 300,000,000 m/s$$

4. Step 4 — Substitute the given values:

   $$E = \Delta m\,300000000^2$$

5. Step 5 — Evaluate:

   $$E = 63900000000000\ \text{J}$$

6. Step 6 — Check: returning E = 63,900,000,000,000 J to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 127,800,000,000,000 — kept a factor of two that cancels in the correct rearrangement.
• 31,950,000,000,000 — dropped that same factor in the other direction.
• 70,290,000,000,000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy

nuclear energy released from a fission mass defect calculation Given speed of light (c) = 300,000,000 m/s; energy released (E) = 4,983,000,000,000 J, determine the mass defect (dm) in kg.

Start with the thinking

• The governing relation printed in this handbook section is Nuclear.
• Everything except dm is given, so isolate dm symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Nuclear energy release in a reactor is described by mass-energy equivalence between mass defect and energy produced.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$E = \Delta m \, c^2$$

2. Step 2 — Rearrange symbolically for dm:

   $$dm = \dfrac{E}{c^2}$$

3. Step 3

   $$List the givens: speed of light (c) = 300,000,000 m/s, energy released (E) = 4,983,000,000,000 J$$

4. Step 4 — Substitute the given values:

   $$dm = \dfrac{4983000000000}{300000000^2}$$

5. Step 5 — Evaluate:

   $$dm = 0.0001\ \text{kg}$$

6. Step 6 — Check: returning dm = 0.0001 kg to

   $$E = \Delta m \, c^2$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0001 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0001 — rounded an intermediate value before the final step.

Reference: FE Handbook — Nuclear Energy