Nuclear Unit Conversion Practice Quiz: Master Physics Units

Grade 12th

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| Attempts: 12 | Questions: 20 | Updated: Mar 11, 2026

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1. If Δm=0.006u, then e≈931×0.006=______ mev.

Concept: quick conversion arithmetic. Multiply by 931. This converts a mass defect in u into energy in mev using the standard factor.

Explanation

Concept: quick conversion arithmetic. Multiply by 931. This converts a mass defect in u into energy in mev using the standard factor.

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About This Quiz

Nuclear Unit Conversion Practice Quiz: Master Physics Units - Quiz

This practice focuses on nuclear unit conversions, assessing skills in converting between different units of measurement relevant to nuclear physics. It helps learners master essential concepts such as activity, decay rates, and energy units, which are critical for understanding nuclear reactions and applications. This knowledge is vital for students and... see moreprofessionals in physics, engineering, and related fields. see less

2. What first name or nickname would you like us to use?

You may optionally provide this to label your report, leaderboard, or certificate.

2. Grade 12 wrap-up: the most useful 'workflow' is to:

Find temperature, then guess energy

Compute Δm, convert to energy with 931 mev/u (or c^2), then (optionally) divide by a

Count electrons, then multiply

Use chemical bond tables

Concept: standard method. That’s the standard mass defect → energy method. It turns a measured mass difference into binding energy, and the per-nucleon step helps compare nuclei fairly.

Explanation

Concept: standard method. That’s the standard mass defect → energy method. It turns a measured mass difference into binding energy, and the per-nucleon step helps compare nuclei fairly.

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3. Even if mass defect is small in u, the equivalent energy in mev can be significant.

True

False

Concept: scale of 1u in energy. 1u corresponds to 931 mev. So even thousandths of a unit of mass can correspond to mev-scale energies.

Explanation

Concept: scale of 1u in energy. 1u corresponds to 931 mev. So even thousandths of a unit of mass can correspond to mev-scale energies.

Submit

4. If you know total binding energy and a, BE per nucleon is:

BE × a

BE / a

A / BE

BE − a

Concept: definition of per-nucleon value. Divide by the nucleon number. This gives the average binding energy per nucleon.

Explanation

Concept: definition of per-nucleon value. Divide by the nucleon number. This gives the average binding energy per nucleon.

Submit

5. A nucleus has Δm=0.030u. Another has Δm=0.024u. Which has larger total binding energy?

Second

First

Equal

Cannot tell

Concept: total binding energy comparison. Larger Δm gives larger binding energy. Since binding energy ≈ 931×Δm (in u), the nucleus with Δm=0.030u has the larger total binding energy.

Explanation

Concept: total binding energy comparison. Larger Δm gives larger binding energy. Since binding energy ≈ 931×Δm (in u), the nucleus with Δm=0.030u has the larger total binding energy.

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6. If two nuclei have the same a, the one with larger Δm is typically more tightly bound.

True

False

Concept: comparing same-size nuclei. Larger defect means more binding energy. With the same number of nucleons, a larger Δm indicates a larger binding energy and thus tighter binding.

Explanation

Concept: comparing same-size nuclei. Larger defect means more binding energy. With the same number of nucleons, a larger Δm indicates a larger binding energy and thus tighter binding.

Submit

7. Useful relations here include:

Δm=(sum of free nucleons)-(nucleus mass)

E=Δmc^2

E("mev")≈931Δm("u")

E=Δm/c^2

Concept: correct formulas. a–c are correct; d is incorrect. You multiply by c^2, not divide, to convert a mass defect into energy.

Explanation

Concept: correct formulas. a–c are correct; d is incorrect. You multiply by c^2, not divide, to convert a mass defect into energy.

Submit

8. If Δm=0.001u, energy is about:

0.0931 mev

0.931 mev

9.31 mev

93.1 mev

Concept: basic conversion. 0.001×931=0.931 mev. This is a direct use of the u-to-mev factor.

Explanation

Concept: basic conversion. 0.001×931=0.931 mev. This is a direct use of the u-to-mev factor.

Submit

9. Mass defect calculations can be done using either u→mev or kg→j methods.

True

False

Concept: equivalent calculation routes. Both are equivalent with correct conversions. Using u with 931 mev/u is just a convenient shortcut compared with SI unit conversions.

Explanation

Concept: equivalent calculation routes. Both are equivalent with correct conversions. Using u with 931 mev/u is just a convenient shortcut compared with SI unit conversions.

Submit

10. If a nucleus has BE ≈ 16 mev and a = 4, BE per nucleon is:

2 mev

4 mev

8 mev

16 mev

Concept: per-nucleon calculation. 16/4 = 4 mev per nucleon. Dividing total binding energy by a gives the average per nucleon.

Explanation

Concept: per-nucleon calculation. 16/4 = 4 mev per nucleon. Dividing total binding energy by a gives the average per nucleon.

Submit

11. If Δm is in atomic mass units (u), the energy equivalent is commonly found using:

E=Δm/931

E("mev")≈931×Δm("u")

E("mev")≈Δm("u")/931

E=Δmc

Concept: converting u to mev. Multiply by 931 mev/u. This converts the mass defect in atomic mass units directly into an energy in mev.

Explanation

Concept: converting u to mev. Multiply by 931 mev/u. This converts the mass defect in atomic mass units directly into an energy in mev.

Submit

12. Which is the best reason nuclear energies are so large?

C^2 is tiny

C^2 is enormous, so small mass differences correspond to big energies

Protons have no charge

Electrons live in the nucleus

Concept: the role of c^2. The factor c^2 is huge. That makes nuclear binding energies large even when the mass defect is only a tiny fraction of the nucleus’s total mass.

Explanation

Concept: the role of c^2. The factor c^2 is huge. That makes nuclear binding energies large even when the mass defect is only a tiny fraction of the nucleus’s total mass.

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13. In SI units, if Δm is in kg, e=Δmc^2 gives energy in joules.

True

False

Concept: SI output units. That’s the SI result. Using kg for mass and (m/s)^2 for c^2 produces energy in joules.

Explanation

Concept: SI output units. That’s the SI result. Using kg for mass and (m/s)^2 for c^2 produces energy in joules.

Submit

14. If Δm increases while a stays the same, BE per nucleon:

Decreases

Increases

Stays the same

Becomes zero

Concept: effect of increasing Δm. More defect means more binding energy. With a fixed, a larger Δm means a larger total binding energy and therefore a larger per-nucleon binding.

Explanation

Concept: effect of increasing Δm. More defect means more binding energy. With a fixed, a larger Δm means a larger total binding energy and therefore a larger per-nucleon binding.

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15. Two nuclei have mass defects 0.018u (a=9) and 0.020u (a=12). Which has higher BE per nucleon?

The second

The first

They are equal

Not enough information

Concept: comparing per-nucleon values. First: 0.002u/nucleon; second: 0.00167u/nucleon. Since BE per nucleon is proportional to Δm/a, the first nucleus has the higher BE per nucleon.

Explanation

Concept: comparing per-nucleon values. First: 0.002u/nucleon; second: 0.00167u/nucleon. Since BE per nucleon is proportional to Δm/a, the first nucleus has the higher BE per nucleon.

Submit

16. Comparing BE per nucleon is generally more informative than comparing total BE for nuclei of different sizes.

True

False

Concept: why per-nucleon matters. 'Per nucleon' normalizes size. Totals naturally grow with the number of nucleons, so the per-nucleon value is better for fair comparisons.

Explanation

Concept: why per-nucleon matters. 'Per nucleon' normalizes size. Totals naturally grow with the number of nucleons, so the per-nucleon value is better for fair comparisons.

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17. If Δm=0.040u and a=20, BE per nucleon is about:

0.093 mev

186.0 mev

1.86 mev

18.6 mev

Concept: multi-step calculation. Total =0.040×931=37.24 mev; per nucleon =37.24/20=1.862 mev. This is the binding energy per nucleon estimate from mass defect.

Explanation

Concept: multi-step calculation. Total =0.040×931=37.24 mev; per nucleon =37.24/20=1.862 mev. This is the binding energy per nucleon estimate from mass defect.

Submit

18. Binding energy per nucleon can be approximated by (931×Δm)/____ in mev.

Concept: per-nucleon binding energy. Divide total energy by nucleon number. First convert Δm to total binding energy, then divide by a to get binding energy per nucleon.

Explanation

Concept: per-nucleon binding energy. Divide total energy by nucleon number. First convert Δm to total binding energy, then divide by a to get binding energy per nucleon.

Submit

19. A nucleus has Δm=0.025u. Its binding energy is closest to:

2.33 mev

0.233 mev

23.3 mev

233 mev

Concept: u→mev multiplication. 0.025×931=23.275 mev. Rounding gives about 23.3 mev.

Explanation

Concept: u→mev multiplication. 0.025×931=23.275 mev. Rounding gives about 23.3 mev.

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20. E=Δmc^2 can be used to compute binding energy from mass defect.

True

False

Concept: mass defect to binding energy. Mass defect corresponds to binding energy via c^2. The energy equivalent of Δm is the binding energy associated with forming the nucleus.

Explanation

Concept: mass defect to binding energy. Mass defect corresponds to binding energy via c^2. The energy equivalent of Δm is the binding energy associated with forming the nucleus.

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