The mass defect of a certain nucleus is found to be $0.03$ amu. Its binding energy is:

27.93

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27.93

keV

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27.93

MeV

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27.93

GeV

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Solution

The correct option is A $27.93$ eV
Mass defect $Δ M = 0.03$ amu
Binding energy $E_{B} = Δ M c^{2} = Δ M \times 931.5$ MeV
$∴ E_{B} = 0.03 \times 931.5$ MeV $= 27.93$ MeV

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The mass defect of a certain nucleus is found to be 0.03 amu. Its binding energy is:

The correct option is A 27.93 eVMass defect ΔM=0.03 amu Binding energy EB=ΔMc2=ΔM×931.5 MeV∴ EB=0.03×931.5 MeV =27.93 MeV

The mass defect of a certain nucleus is found to be $0.03$ amu. Its binding energy is:

The correct option is A $27.93$ eV
Mass defect $Δ M = 0.03$ amu
Binding energy $E_{B} = Δ M c^{2} = Δ M \times 931.5$ MeV
$∴ E_{B} = 0.03 \times 931.5$ MeV $= 27.93$ MeV