Question

Given the following binding energies:
$1.{}_8^{17}O\to 131MeV$
$2.{}_{26}^{56}Fe\to 493MeV$
$3.{}_{92}^{238}U\to 1804MeV$
Compare the binding energies per nucleon $(\Delta =\frac{B}{A})$ of the three nuclei.

• A. ${\Delta }_O$>${\Delta }_{Fe}$>${\Delta }_U$
• B. ${\Delta }_{Fe}$>${\Delta }_O$>${\Delta }_U$
• C. ${\Delta }_U$>${\Delta }_O$>${\Delta }_{Fe}$
• D. ${\Delta }_U$>${\Delta }_{Fe}$>${\Delta }_O$

Answer 1

The correct option is B ${\Delta }_{Fe}$>${\Delta }_O$>${\Delta }_U$

The binding energy per nucleon is just the average binding energy shared by each nucleon, given as,
$\frac{Netbindingenergy}{Numberofnucleons}=\frac{Netbindingenergy,B}{Massnumber,A}.$
The upper left number, i , in the representation ${}_j^{i}X$ tells us the mass number, or total number of nucleons in the nucleus. Therefore, for ${}_8^{17}O,A=17.For{}_{26}^{56}Fe,A=56.Andfor{}_{92}^{238}U,A=238$ ${\left[\frac{B}{A}\right]}_{{17}_O}=\frac{131}{17}MeV=7.7MeV,$
${\left[\frac{B}{A}\right]}_{{56}_{Fe}}=\frac{493}{56}MeV=8.8MeV,$
${\left[\frac{B}{A}\right]}_{{238}_U}=\frac{1804}{238}MeV=7.5MeV$
Therefore, representing the binding energy by $\Delta$,
${\Delta }_{Fe}$>${\Delta }_O$>${\Delta }_U$