The neutral ${ } _ { 13 } ^ { 27 } \mathrm { Al }$ atom has a mass of $26.981539$ u; a neutral hydrogen atom has a mass of $1.007825$ u; a neutron has a mass of $1.008665 \mathrm { u }$; and a proton has a mass of $1.007277 \mathrm { u }$. What is the binding energy per nucleon for ${ } _ { 13 } ^ { 27 } \mathrm { Al }$ ? $\left( 1 \mathrm { u } = 931.5 \mathrm { MeV } / c ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

The binding energy per nucleon is given by the formula:$$B/A = (Zm_p + Nm_n - M)/A$$where Z is the atomic number, N is the neutron number, m_p is the mass of a proton, m_n is the mass of a neutron, M is the mass of the nucleus, and A is the mass number.For ${ } _ { 13 } ^ { 27 } \mathrm { Al }$, Z = 13, N = 14, m_p = 1.007277 u, m_n = 1.008665 u, M = 26.981539 u, and A = 27.Plugging these values into the formula, we get:$$B/A = (13 \times 1.007277 \mathrm { u } + 14 \times 1.008665 \mathrm { u } - 26.981539 \mathrm { u })/27 = 8.3 \mathrm { MeV }$$Therefore, the binding energy per nucleon for ${ } _ { 13 } ^ { 27 } \mathrm { Al }$ is 8.3 MeV.

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The Neutral ${ } _ { 13 } ^ { 27 } \mathrm { Al }$