Mass defect of tritium

Problem

Tritium $^{3}_{1}\mathrm{H}$ has atomic mass $3{,}016049\ \text{u}$. It is made up of 1 proton ($1{,}007276\ \text{u}$), 2 neutrons ($1{,}008665\ \text{u}$ each) and 1 electron ($0{,}000549\ \text{u}$). Calculate: (a) the mass defect, (b) the binding energy in MeV, (c) the binding energy per nucleon.

Solution

(a) Mass defect. The mass defect $\Delta m$ is the difference between the sum of the masses of the constituents and the actual mass of the nucleus (atom). First we sum the masses of the constituents:

$m_{\text{constituents}} = 1{,}007276 + 2\cdot 1{,}008665 + 0{,}000549 = 3{,}025155\ \text{u}$

The mass defect is then:

$\Delta m = m_{\text{constituents}} - m_{\text{atom}} = 3{,}025155 - 3{,}016049 = \ev{0{,}009106\ \text{u}}$

(b) Binding energy. Using the equivalence $1\ \text{u} \leftrightarrow 931{,}5\ \text{MeV}$:

$B = \Delta m \cdot 931{,}5\ \tfrac{\text{MeV}}{\text{u}} = 0{,}009106 \cdot 931{,}5 \approx \ev{8{,}48\ \text{MeV}}$

(c) Binding energy per nucleon. The number of nucleons is $A = 3$ (1 proton + 2 neutrons):

$\frac{B}{A} = \frac{8{,}48}{3} \approx \ev{2{,}83\ \tfrac{\text{MeV}}{\text{nucleon}}}$

This value is low compared with the $\sim 8\ \text{MeV/nucleon}$ typical of stable mid-mass nuclei: tritium is a light, weakly bound nucleus, consistent with its instability (it undergoes $\beta^-$ decay).

Collegamenti

Argomenti: Fisica nucleare Concetti: Energia di legame nucleare · Equivalenza massa-energia