Question

Natural nitrogen is 99.63% ${}^{14}N$ with mass 14.00307 amu and 0.37% ${}^{15}N$, with mass number 15.00011 amu. Which isotope is more stable, ${}^{14}N$ or ${}^{15}N$ ?
Plan Binding energy per nucleon$=\frac{△m\left(amu\right)\times 931}{numberofnucleons}$
Isotope with larger binding energy per nucleon is more stable.

• A. $7.695$ MeV
• B. $2.995$ MeV
• C. $8.8695$ MeV
• D. None of these

Answer 1

The correct option is A $7.695$ MeV

There are 7n, 7p and $7e^{-}$ in ${}^{14}N$
mass of nucleons$=7\times 1.00782505+7\times 1.008664904$
$=14.11543$ amu
mass defect $=14.11543-14.00307$
$=0.11236$ amu
binding energy per nucleon $=\frac{0.11236\times 931}{14}$ MeV
$=7.472$ MeV
There are 8n, 7p and $7e^{-}$ in ${}^N$
mass of nucleons$=8\times 1.008664904+7\times 1.00782505$
$=15.12409$ amu
mass defect $=15.12409-15.00011$
$=0.12398$ amu
binding energy per nucleon$=\frac{0.12398\times 931}{15}MeV$
$7.695$ MeV
As binding energy per nucleon is larger for ${}^{15}N$, hence, ${}^{15}N$ is more stable.