Question 70: Atoms & Nuclei

Question

The energy released if hydrogen atoms are combined to form ${ }_2^4 \mathrm{He}$ is $\_\_\_\_$ MeV .
(Take binding energies per nucleon of ${ }_1^2 \mathrm{H}$ and ${ }_2^4 \mathrm{He}$ as 1.1 MeV and 7.2 MeV , respectively)

Accepted Answer

The binding energy (B.E.) of a nucleus is the energy required to disassemble it into its constituent nucleons.
Total B.E. $=($ Binding Energy per nucleon $) 	imes($ Number of nucleons A$)$
Energy Released (Q) = (Total B.E. of Products) - (Total B.E. of Reactants)
The given reaction is $\mathrm{H}_1^2+\mathrm{H}_1^2 ightarrow \mathrm{He}_2^4$
The reactants are two nuclei of $\mathrm{H}_1^2$.
B.E. per nucleon of $\mathrm{H}_1^2=1.1 \mathrm{MeV}$
Number of nucleons (A) for one $\mathrm{H}_1^2=2$
So, total B.E. of reactants $=2 	imes(1.1 	imes 2)=4.4 \mathrm{MeV}$
The product is one nucleus of $\mathrm{He}_2^4$.
B.E. per nucleon of $\mathrm{He}_2^4=7.2 \mathrm{MeV}$
Number of nucleons (A) for $\mathrm{He}_2^4=4$
So, total B.E. of product $=7.2 	imes 4=28.8 \mathrm{MeV}$
Hence, energy released during reaction is,
$$ \mathrm{Q}=28.8 \mathrm{MeV}-4.4 \mathrm{MeV} $$
$$\Rightarrow $$ $$ \mathrm{Q}=24.4 \mathrm{MeV} $$
Therefore, the energy released if hydrogen atoms are combined to form $\mathrm{He}_2^4$ is 24.4 MeV .
Thus, the correct option is (D).

Answer

6.1

Answer

24.4

Answer

26.6

Answer

5

Atoms & Nuclei

The energy released if hydrogen atoms are combined to form { }_2^4 \mathrm{He} is \_\_\_\_ MeV . (Take binding energies per nucleon of { }_1^2 \mathrm{H} and { }_2^4 \mathrm{He} as 1.1 MeV and 7.2 MeV , respectively)

Choose the correct answer: