Question:medium

The binding energy per nucleon of \(^{209} \text{Bi}\)  is _______ MeV. \[ \text{Take } m(^{209} \text{Bi}) = 208.98038 \, \text{u}, \, m_p = 1.007825 \, \text{u}, \, m_n = 1.008665 \, \text{u}, \, 1 \, \text{u} = 931 \, \text{MeV}/c^2. \]

Updated On: Apr 10, 2026
  • 7.48
  • 7.84
  • 8.79
  • 6.94
Show Solution

The Correct Option is D

Solution and Explanation

Was this answer helpful?
0


Questions Asked in JEE Main exam