Atoms And Nuclei Question 239

Question: The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form helium atom, the energy released is [Pb. PMT 2001; CPMT 2001; AIEEE 2004]

Options:

A) 19.2 MeV

B) 23.6 MeV

C) 26.9 MeV

D) 13.9 MeV

Show Answer

Answer:

Correct Answer: B

Solution:

$ _{1}H^{2}+ _{1}H^{2}\to _{2}He^{4}+ $ energy

Binding energy of a $ _{1}H^{2} $ deuterium nuclei '

$ =2\times 1.1=2.2\ MeV $ '

Total binding energy of two deuterium nuclei $ =2.2\times 2=4.4\ MeV $ B'

inding energy of a $ _{2}He^{4} $ nuclei $ =4\times 7=28\ MeV $ '

So, energy released in fusion $ =28-4.4=23.6\ MeV $



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