KCET · Physics · Current Electricity
The mass defect of \( { }_{2}^{4} \mathrm{He} \) is \( 0.03 \mathrm{u} \). The binding energy per nucleon of helium (in \( \mathrm{MeV} \) ) is
- A \( 27.93 \)
- B \( 6.9825 \)
- C \( 2.793 \)
- D \( 69.825 \)
Answer & Solution
Correct Answer
(B) \( 6.9825 \)
Step-by-step Solution
Detailed explanation
Binding energy is the amount of energy required to separate a particle from a system of particles. It is given as
\( B=\frac{(\Delta m) c^{2}}{A} \)
where \( \Delta \mathrm{m} \) is the mass defect
Given, mass defect \( \Delta \mathrm{m}=0.03 \mathrm{u}, \mathrm{A}=4 \)
Therefore, binding energy per nucleon of helium is
\( B=\frac{0.03 \times 931}{4} \mathrm{MeV}=6.9825 \mathrm{MeV} \)
\( B=\frac{(\Delta m) c^{2}}{A} \)
where \( \Delta \mathrm{m} \) is the mass defect
Given, mass defect \( \Delta \mathrm{m}=0.03 \mathrm{u}, \mathrm{A}=4 \)
Therefore, binding energy per nucleon of helium is
\( B=\frac{0.03 \times 931}{4} \mathrm{MeV}=6.9825 \mathrm{MeV} \)
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