CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 14 of 14

Marks: +1, -0

(a) Calculate the binding energy of an alpha particle in MeV. Given
mass of a proton

=1.007825 {\u}

mass of a neutron

=1.008665 {\u}

mas of He nucleus

=4.002800 u

1 {\u}=931 {\MeV} \/ {\c}^2

OR
(b) A heavy nucleus

P

of mass number 240 and binding energy 7.6

{\MeV}

per nucleon splits into two nuclei

Q

and

R

of mass number 110 and 130 and binding energy per nucleon

8.5 {\MeV}

and

8.4 {\MeV}

respectively. Calculate the energy released in the fission.

Solution:

(a) Mass defect

=

Mass of protons + mass of neutrons - Mass of Helium nucleus

\;\text " Or, "\; \;\; \;\text " Mass defect "\;=\;(2 × 1.007825+2

\;× 1.008665-4.002800) {\u}

∴ \;\; \;\text " Mass defect "\;=\;0.03018 {\u}

=\;0.03018 × 931

=\;28.09785 {\MeV}

OR
(b) According to the question

{ }^{240} {\P} →{ }^{110} {\Q}+{ }^{130} R+Q

Q=\;110 × 8.5+130 × 8.4

\;-240 × 7.6

Or,

Q\;=935+1092-1824

∴

Q\;=202 {\MeV}

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