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Atoms and Nuclei question

2009 · Shift 0 · Q50
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Atoms and Nuclei question

2009 · Shift 0 · Q50

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
AIEEE 2009 Physics - Atoms and Nuclei Question 229 English The above is a plot of binding energy per nucleon Eb,{E_b},Eb​, against the nuclear mass M;A,B,C,D,E,FM;A,B,C,D,E,FM;A,B,C,D,E,F correspond to different nuclei. Consider four reactions : (i)          A+B→C+ε          (ii)          C→A+B+ε          (iii)      D+E→F+ε          (iv)         F→D+E+ε,          \begin{aligned} & \left( i \right)\,\,\,\,\,\,\,\,\,\,A + B \to C + \varepsilon \,\,\,\,\,\,\,\,\,\,\left( {ii} \right)\,\,\,\,\,\,\,\,\,\,C \to A + B + \varepsilon \,\,\,\,\,\,\,\,\,\, \\ & \left( {iii} \right)\,\,\,\,\,\,D + E \to F + \varepsilon \,\,\,\,\,\,\,\,\,\,\left( {iv} \right)\,\,\,\,\,\,\,\,\,F \to D + E + \varepsilon ,\,\,\,\,\,\,\,\,\,\, \\\end{aligned}​(i)A+B→C+ε(ii)C→A+B+ε(iii)D+E→F+ε(iv)F→D+E+ε,​ where ε\varepsilonε is the energy released? In which reactions is ε\varepsilonε positive?
  1. A
    (i)(i)(i) and (iii)(iii)(iii)
  2. B
    (ii)(ii)(ii) and (iv)(iv)(iv)
  3. C
    (ii)(ii)(ii) and (iii)(iii)(iii)
  4. D
    (i)(i)(i) and (iv)(iv)(iv)
View written solutionFree

Correct answer: D

  1. Key physical principle

For a nuclear reaction, energy is released when the total binding energy increases.

If the products have larger binding energy than the reactants, then

ε>0.\varepsilon > 0.ε>0.

Since binding energy per nucleon is plotted, the total binding energy is approximately

B=A (Eb),B = A\,(E_b),B=A(Eb​),

where AAA is mass number.

Also, from the standard binding energy per nucleon curve:

  • Light nuclei gain energy by fusion if they move toward higher EbE_bEb​.
  • Very heavy nuclei gain energy by fission if they split into nuclei with higher EbE_bEb​.

  1. Interpretation of the graph

Although the figure is not shown here, this standard question uses the usual binding-energy-per-nucleon curve:

  • A,B,CA,B,CA,B,C lie in the lighter side region where combining lighter nuclei can move to a nucleus with larger EbE_bEb​.
  • D,E,FD,E,FD,E,F lie in the heavier side region where splitting a heavy nucleus can move to nuclei with larger EbE_bEb​.

So we test each reaction qualitatively.


  1. Reaction (i):
A+B→C+εA+B \to C + \varepsilonA+B→C+ε

This is fusion of lighter nuclei to form a heavier nucleus CCC.

For light nuclei, fusion releases energy if the product has higher binding energy per nucleon than the reactants. From the binding-energy curve, this is true.

Hence,

ε>0\varepsilon > 0ε>0

for (i).


  1. Reaction (ii):
C→A+B+εC \to A+B+\varepsilonC→A+B+ε

This is the reverse of reaction (i). If (i) releases energy, then the reverse must require energy input.

So for (ii), energy is not released:

ε<0(or equivalently, energy must be supplied).\varepsilon < 0 \quad \text{(or equivalently, energy must be supplied).}ε<0(or equivalently, energy must be supplied).

Thus (ii) is not positive.


  1. Reaction (iii):
D+E→F+εD+E \to F+\varepsilonD+E→F+ε

This is fusion on the heavy side of the binding-energy curve.

For heavy nuclei, fusion generally moves to lower binding energy per nucleon, so it does not release energy.

Hence for (iii),

ε<0\varepsilon < 0ε<0

so (iii) is not positive.


  1. Reaction (iv):
F→D+E+εF \to D+E+\varepsilonF→D+E+ε

This is fission of a heavy nucleus. On the heavy side, splitting into medium-mass nuclei increases binding energy per nucleon, so energy is released.

Therefore,

ε>0\varepsilon > 0ε>0

for (iv).


  1. Final selection

The reactions with positive energy release are:

(i) and (iv)(i) \text{ and } (iv)(i) and (iv)

So the correct option is

D\boxed{\text{D}}D​
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