Given the following binding energies-1- 817O - 131 MeV2- 5626Fe - 493 MeV3- 92238U - 1804 MeVCompare the binding energies per nucleon -B-A of the three nuclei-

The binding energy per nucleon is just the average binding energy shared by each nucleon, given as, Net binding energy/Number of nucleons=Net binding energy, B/ ...

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Standard XII

Chemistry

Binding Energy

Given the fol...

Question

Given the following binding energies:
$1._{8}^{17} O \to 131 M e V$
$2._{26}^{56} F e \to 493 M e V$
$3._{92}^{238} U \to 1804 M e V$
Compare the binding energies per nucleon $(Δ = \frac{B}{A})$ of the three nuclei.

Δ_{O}

Δ_{F e}

Δ_{U}

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Δ_{F e}

Δ_{O}

Δ_{U}

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Δ_{U}

Δ_{O}

Δ_{F e}

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Δ_{U}

Δ_{F e}

Δ_{O}

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Solution

The correct option is B $Δ_{F e}$ > $Δ_{O}$ > $Δ_{U}$
The binding energy per nucleon is just the average binding energy shared by each nucleon, given as,
$\frac{N e t b i n d i n g e n e r g y}{N u m b e r o f n u c l e o n s} = \frac{N e t b i n d i n g e n e r g y, B}{M a s s n u m b e r, A} .$
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 $_{8}^{17} O, A = 17. F o r_{26}^{56} F e, A = 56. A n d f o r_{92}^{238} U, A = 238$ ${[\frac{B}{A}]}_{17_{O}} = \frac{131}{17} M e V = 7.7 M e V,$
${[\frac{B}{A}]}_{56_{F e}} = \frac{493}{56} M e V = 8.8 M e V,$
${[\frac{B}{A}]}_{238_{U}} = \frac{1804}{238} M e V = 7.5 M e V$
Therefore, representing the binding energy by $Δ$ ,
$Δ_{F e}$ > $Δ_{O}$ > $Δ_{U}$

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Given the following binding energies: 1. 178O→131 MeV 2. 5626Fe→493 MeV 3. 23892U→1804 MeV Compare the binding energies per nucleon (Δ=BA) of the three nuclei.

Given the following binding energies:
$1._{8}^{17} O \to 131 M e V$
$2._{26}^{56} F e \to 493 M e V$
$3._{92}^{238} U \to 1804 M e V$
Compare the binding energies per nucleon $(Δ = \frac{B}{A})$ of the three nuclei.