Binding energy Confusion

[tRNA]

the nuclear binding energy, EB, can be found from the mass defect using Einstein's equation for mass-energy equivalence: EB=delta mc^2, where c is the speed or light (3x10^8m/s). if mass is measured in kilograms and energy in joules, then 1kg <--> 9x10^16J. But in the nuclear domain, masses are often expressed in atomic mass units(1amu=1.66x10^(-27)kg), and energy is expressed in electronvolts (1eV=1.6x10^(-19)J). In terms of these units, the equation for the nuclear binding energy, EB=deltamc^2, can be written as EB(in eV)= delta m (in amu)x 931 MeV

the above paragraph was very confusing to me, can someone Please explain it and how does it relate to the below problem, and why do they ignore the speed of light from the EB equation?:

1) the mass defect of helium nucleus is 5x10^(-29)Kg what is the nuclear binding energy?

my answer:
since EB=delta mc^2
then EB= 5x10^(-29)Kg ( 3x10^8m/s)= 1.5x10^(-20)

the correct answer:
the equation EB=delta mc^2 implies that 1kg <--> 9x10^16J, so a mass defect of 5x10^(-29)Kg is equivalent to an energy of (5x10^(-29)Kg) ( 9x10^16J)=4.5x10^(-12)J

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[tRNA]

can someone please help???,what am I doing wrong?? do we get binding energy problems like this one on the test?

 

[MDwannabe7]

the nuclear binding energy, EB, can be found from the mass defect using Einstein's equation for mass-energy equivalence: EB=delta mc^2, where c is the speed or light (3x10^8m/s). if mass is measured in kilograms and energy in joules, then 1kg <--> 9x10^16J. But in the nuclear domain, masses are often expressed in atomic mass units(1amu=1.66x10^(-27)kg), and energy is expressed in electronvolts (1eV=1.6x10^(-19)J). In terms of these units, the equation for the nuclear binding energy, EB=deltamc^2, can be written as EB(in eV)= delta m (in amu)x 931 MeV

the above paragraph was very confusing to me, can someone Please explain it and how does it relate to the below problem, and why do they ignore the speed of light from the EB equation?:

1) the mass defect of helium nucleus is 5x10^(-29)Kg what is the nuclear binding energy?

my answer:
since EB=delta mc^2
then EB= 5x10^(-29)Kg ( 3x10^8m/s)= 1.5x10^(-20)

the correct answer:
the equation EB=delta mc^2 implies that 1kg <--> 9x10^16J, so a mass defect of 5x10^(-29)Kg is equivalent to an energy of (5x10^(-29)Kg) ( 9x10^16J)=4.5x10^(-12)J

The first mistake is that you need to square the speed of light - that actually gives 4.5x10^-12 J, which is the correct answer.

To explain how

the equation for the nuclear binding energy, EB=deltamc^2, can be written as EB(in eV)= delta m (in amu)x 931 MeV

, I just figured this out myself, so I'll try to explain it to you.

The atomic mass unit, amu, can be defined as 1/12 the mass of a carbon 12 atom, which is equal to 1.660540x10^-27 kg. Using Einstein's Theory of Relativity, we can determine the energy equivalent of one amu.

E=mc^2; m of 1 amu = 1.660540x10^-27 kg; c = 3.00x10^8 m/s

Plugging these values in gives us the energy equivalent in Joules (a Joule = Newton*Meter = kg*m^2/s^2) = 1.494486x10^-10 J.

Since physicists like to work in eV, 1 eV = 1.6022x10^-19 J. Therefore, if we divide our answer in J by the conversion factor, we get 9.32x10^8 or 932 MeV. Therefore, the energy equivalent of 1 amu is 932 MeV, which can be plugged into the equation above my multiplying your number of amus by 932 MeV.

5x10^-29 kg * 1 amu/1.660540x10^-27 kg * 932x10^6 eV/1 amu * 1.6022x10^-19/1 eV = 4.496x10^-12 J

In their correct answer explanation, they state

the equation EB=delta mc^2 implies that 1kg <--> 9x10^16J

That simply comes from combining the three terms mentioned above (1 amu/1.660540x10^-27 kg * 932x10^6 eV/1 amu * 1.6022x10^-19/1 eV = 9x10^16 J)

Hope this helps - I finally get it.