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The atomic masses of Li, He and proton a...

The atomic masses of Li, He and proton are 7.01823 amu, 4.00387 amu and 1.00715 amu respectively. Calculate the energy evolved in the reaction, `3^(Li^(7)) + 1^(HA^(1)) rarr 2 2^(He^(4)) + triangle` E Given 1 amu = 931 MeV.

Text Solution

Verified by Experts

Mass of reactants = mass of Li + mass of H
=7.01823 + 1.00715
=8.02358 amu
mass loss during change =(8.02538 -8.00774) amu
=0.01764 amu
`therefore` energy evolved during reaction
=0.0176 x 931 MeV
=16.423 MeV
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If the atomic masses of lithium, helium and proton are 7.01823 amu, 4.00387 amu and 1.00815 amu respectively, calculate the energy that will be evolved in the reaction, Li^7 + H^1 + 2He^4 + energy Given that 1 amu 931 MeV.

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Knowledge Check

  • The binding energy per nucleon of ._3^7Li and ._2^4He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction ._3^7Li + ._1^1H to 2._2^4He + Q , the value of energy Q released is

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