use the unified atomic mass unit (u) as a unit of mass

Objective

Students will be able to use the unified atomic mass unit (u) as a unit of mass in calculations involving atoms, nuclei and radiation.

1. The Unified Atomic Mass Unit

1.1 Definition

The unified atomic mass unit, symbolised by u (also written as amu), is defined as one twelfth of the mass of a neutral carbon‑12 atom:

$$ 1\ \text{u} = \frac{1}{12}m_{\text{C}^{12}} $$

1.2 \cdot alue in kilograms

Experimentally, the value of 1 u in SI units is:

$$ 1\ \text{u}=1.66053906660\times10^{-27}\ \text{kg} $$

This conversion factor allows us to move between the atomic scale and the macroscopic scale.

1.3 Why use the unified atomic mass unit?

• Atomic and nuclear masses are of the order of 10⁻²⁷ kg, making the kilogram inconvenient.
• Using u provides a convenient scale that directly relates to the number of nucleons (protons + neutrons) in a nucleus.
• It simplifies stoichiometric calculations in chemistry and particle‑physics problems.

2. Relating Mass Number to Mass in u

2.1 Mass number (A)

The mass number A of an atom is the total number of protons and neutrons in its nucleus. For most nuclides the mass in unified atomic mass units is close to A:

$$ m \approx A\ \text{u} $$

Small deviations arise from binding energy differences (mass defect).

2.2 Example: Carbon‑12

Carbon‑12 has A = 12. Its mass is defined as exactly 12 u, which corresponds to:

$$ 12\ \text{u}=12\times1.66053906660\times10^{-27}\ \text{kg}=1.9926468799\times10^{-26}\ \text{kg} $$

3. Converting Between u and kg

3.1 General conversion formula

To convert a mass m given in unified atomic mass units to kilograms:

$$ m\ (\text{kg}) = m\ (\text{u}) \times 1.66053906660\times10^{-27}\ \text{kg/u} $$

3.2 Practice conversion

1. Find the mass in kilograms of a uranium‑235 nucleus (A = 235).
2. Express the mass of a neutron (≈ 1.008 u) in kilograms.

4. Applications in Radiation Physics

4.1 Energy–mass equivalence

Using Einstein’s relation $E = mc^{2}$, the mass of a particle expressed in u can be converted directly to energy in mega‑electron‑volts (MeV):

$$ E\ (\text{MeV}) = m\ (\text{u}) \times 931.494\ \text{MeV/u} $$

Here $c$ is the speed of light in vacuum, $c = 2.998\times10^{8}\ \text{m s}^{-1}$.

4.2 Example: Alpha decay of $^{238}\text{U}$

The reaction is:

$$ ^{238}\text{U} \rightarrow\ ^{234}\text{Th} +\ ^{4}\text{He} + Q $$

Using atomic masses (in u):

• $m(^{238}\text{U}) = 238.050788\ \text{u}$
• $m(^{234}\text{Th}) = 234.043601\ \text{u}$
• $m(^{4}\text{He}) = 4.002603\ \text{u}$

The Q‑value (energy released) is:

$$ Q = \bigl[m(^{238}\text{U}) - m(^{234}\text{Th}) - m(^{4}\text{He})\bigr]c^{2} = (0.004584\ \text{u})\times931.494\ \text{MeV/u} \approx 4.27\ \text{MeV} $$

5. Summary Table of Common Nuclides

Nuclide	Mass number (A)	Atomic mass (u)	Mass (kg)
$^{1}\text{H}$ (protium)	1	1.007825	1.674 × 10⁻²⁷
$^{12}\text{C}$	12	12.000000	1.992 × 10⁻²⁶
$^{14}\text{N}$	14	14.003074	2.324 × 10⁻²⁶
$^{238}\text{U}$	238	238.050788	3.953 × 10⁻²⁵

6. Practice Questions

1. Calculate the mass in kilograms of a single $^{56}\text{Fe}$ nucleus (atomic mass = 55.9349 u).
2. A beta particle has a kinetic energy of 0.511 MeV. Using $E = mc^{2}$, find its mass in u.
3. Determine the Q‑value for the beta decay $^{14}\text{C} \rightarrow\ ^{14}\text{N} + e^{-} + \bar{u}_{e}$ given:
   • $m(^{14}\text{C}) = 14.003242\ \text{u}$
   • $m(^{14}\text{N}) = 14.003074\ \text{u}$
   • $m(e^{-}) = 0.0005486\ \text{u}$

7. Suggested Diagram