The binding energies per nucleon are $$5.3$$ MeV, $$6.2$$ MeV and $$7.4$$ MeV for the nuclei with mass numbers $$3$$, $$4$$ and $$5$$ respectively. If one nucleus of mass number $$3$$ combines with one nucleus of mass number $$5$$ to give two nuclei with mass number $$4$$. Then

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$$Bi=3$$  ×  $$5.3+5$$  ×  $$7.4$$ $$MeV=52.9$$  MeV  $$Bf=$$  $$4$$  ×  $$6.2$$ ×  $$2$$    MeV  $$=$$  $$49.6$$  MeVSince Bi > Bf, energy is absorbed∴Energy absorbed $$=$$ Bi  −   $$Bf=3.3$$  MeV

Nuclear Binding energy

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The binding energies per nucleon are 5.3 MeV, 6.2 MeV and 7.4 MeV for the nuclei with mass numbers 3, 4 and 5 respectively. If one nucleus of mass number 3 combines with one nucleus of mass number 5 to give two nuclei with mass number 4. Then

$\begin{array}{l} B_{i} = (3 \times 5 .3 + 5 \times 7 .4) MeV = 52 .9 MeV \\ B_{f} = (4 \times 6 .2) \times 2 MeV = 49 .6 MeV \end{array}$
Since $B_{i} > B_{f},$ energy is absorbed
$∴$ Energy absorbed = $B_{i} - B_{f} = 3 .3 MeV$

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Nuclear Binding energy

Remember concepts with our Masterclasses.

The binding energies per nucleon are 5.3 MeV, 6.2 MeV and 7.4 MeV for the nuclei with mass numbers 3, 4 and 5 respectively. If one nucleus of mass number 3 combines with one nucleus of mass number 5 to give two nuclei with mass number 4. Then

Bi=3 × 5.3+5 × 7.4 MeV=52.9 MeV Bf= 4 × 6.2 × 2 MeV = 49.6 MeVSince Bi > Bf, energy is absorbed∴Energy absorbed = Bi − Bf=3.3 MeV

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$\begin{array}{l} B_{i} = (3 \times 5 .3 + 5 \times 7 .4) MeV = 52 .9 MeV \\ B_{f} = (4 \times 6 .2) \times 2 MeV = 49 .6 MeV \end{array}$
Since $B_{i} > B_{f},$ energy is absorbed
$∴$ Energy absorbed = $B_{i} - B_{f} = 3 .3 MeV$