Modern Physics 5 Question 5
5. Consider the nuclear fission
$Ne^{20} \rightarrow 2 He^{4}+C^{12}$
Given that the binding energy/nucleon of $Ne^{20}, He^{4}$ and $C^{12}$ are respectively, $8.03 MeV, 7.07 MeV$ and $7.86 MeV$, identify the correct statement.
(Main 2019, 10 Jan II)
(a) Energy of $3.6 MeV$ will be released.
(b) Energy of $12.4 MeV$ will be supplied.
(c) $8.3 MeV$ energy will be released.
(d) Energy of $11.9 MeV$ has to be supplied.
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Answer:
Correct Answer: 5. (*)
Solution:
- Energy absorbed or released in a nuclear reaction is given by
$\Delta Q=$ Binding energy of products
-Binding energy of reactants.
If energy is absorbed, $\Delta Q$ is negative and if it is positive then energy is released. Also, Binding energy $=$ Binding energy per nucleon $\times$ Number of nucleons.
Here, binding energy of products
$$ \begin{aligned} & =2 \times\left(\text { B.E. of } He^{4}\right)+\left(\text { B.E. of } C^{12}\right) \\ & =2(4 \times 7.07)+(12 \times 7.86)=150.88 MeV \end{aligned} $$
and binding energy of reactants $=20 \times 8.03=160.6 MeV$
So, $\begin{aligned} \Delta Q= & (\text { B.E. }) _{\text {Products }}-(\text { B.E. }) _{\text {reactants }} \ & =150.88-160.6=-9.72 MeV\end{aligned}$
As $\Delta Q$ is negative
$\therefore$ energy of $9.72 Mev$ is absorbed in the reaction.
$\therefore$ No option is correct
