CBSE Class 12 Physics 2016 Delhi Set 1 Paper

Question : 8 of 26

Marks: +1, -0

A nucleus with mass number

A=240

and

B E \/ A=7.6 {\MeV}

breaks into two fragments each of

A=120

with

B E \/ A=8.5 {\MeV}

. Calculate the released energy.
OR
Calculate the energy in fusion reaction :

{ }_1^2 {\H}+{ }_1^2 {\H} →{ }_2^3 {\He}+ {\n}

, where BE of

{ }_1^2 {\H}=2.23 {\MeV}

and of

{ }_2^3 {\He}=7.73 {\MeV}

Solution:

Calculation of energy released
Binding energy of nucleus with mass number 240 ,

E_{b n}=240 × 7.6 {\MeV}

Binding energy of two fragments

\;=2 × 120 × 8.5 {\MeV}

\;\text " Energy released "\;=\;240(8.5-7.6) {\MeV}

\;=240 × 0.9

\;=216 {\MeV}

OR
Calculation of Energy in the fusion Reaction
Total Binding energy of Initial System i.e.

{ }_1^2 {\H}+{ }_1^2 {\H}\;=(2.23+2.23) {\MeV}

\;=4.46 {\MeV}

Binding energy of Final System i.e.

{ }_2^3 {\He}

=7.73 {\MeV}

Hence energy released

=7.73 {\MeV}-4.46 {\MeV}

=3.27 {\MeV}

Go to Question:

Prev Question

Next Question