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An atom of lithium-6 has a mass of 6.01513 amu. The sum of the masses of this isotope's nucleons is 6.0497 amu. Calculate the binding energy in MeV. Remember 1 amu = 931.5 MeV. Enter your answer using three significant figures. Answer:___ MeV

**nvm, use quizlet ya'll : here https://quizlet.com/380580726/final-exam-review-chem-flash-cards/ ** it's 32.2 MeV btw

Respuesta :

ardni313

The binding energy  : 32.2 MeV

Further explanation

The binding energy : the energy to split the nucleus into individual protons and neutrons.

Can be formulated :

[tex]\tt (mass~of~nucleons-mass~of~atom)\times 931.5~MeV[/tex]

So, add the masses of subatomic particles of atom(proton+electron+neutron) and then substrac the mass of atom

Mass of subatomic particles :

  • neutron : 1.008664 u
  • proton : 1.007276 u
  • electron : 0.00054858 u

For lithium-6 there are 3 protons, 3 electrons and 3 neutron

we can prove the sum of the masses of this isotope's nucleons of Lithium-6 above (which is 6.0497 amu.)

=3 x  1.008664 + 3 x 1.007276 + 3 x 0.00054858

=3.025995 + 3.021828+0.00164574=6.0495 amu(close to the value in the problem)

okay let's finish the question

The binding energy in MeV.

[tex]\tt (6.0497-6.01513)\times 931.5~MeV=\boxed{\bold{32.202~MeV}}[/tex]

to 3 significant figure ⇒ 32.2 MeV

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