The binding energy per nucleon of 209 83Bi is _______ MeV. [Take m(209 83Bi) = 208.980388 u, mp = 1.007825 u, mn = 1.008665 u, 1 u=931 MeV/c^2]

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q44 – JEE-MAIN 2026

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Home Physics NucleiQuestion #365

The binding energy per nucleon of

_{83}^{209} Bi

is _______ MeV. \ [Take

m (_{83}^{209} Bi) = 208.980388 u

m_{p} = 1.007825 u

m_{n} = 1.008665 u

1 u = 931 {MeV/c}^{2}

]

Clue 1 of 3: Nuclear Structure

Understand that the binding energy arises from the mass defect, which is the difference between the mass of individual nucleons and the mass of the nucleus.

🧠 Full Solution Path

Step 1: Determine the number of protons and neutrons✦ Active

For the nucleus $_{83}^{209} Bi$ , the atomic number $Z = 83$ (number of protons) and the mass number $A = 209$ (total number of nucleons). The number of neutrons $N = A - Z = 209 - 83 = 126$ .

Step 2: Calculate the mass defect○ Expand

The total mass of individual nucleons is $M_{nucleons} = Z \cdot m_{p} + N \cdot m_{n}$ .

M_{nucleons} = (83 \times 1.007825 u) + (126 \times 1.008665 u)

M_{nucleons} = 83.649475 u + 127.09179 u = 210.741265 u

The mass defect is $Δ m = M_{nucleons} - M_{Bi}$ .

Δ m = 210.741265 u - 208.980388 u = 1.760877 u

Step 3: Calculate the binding energy per nucleon○ Expand

The total binding energy is $B E = Δ m \times 931 MeV/u$ .

B E = 1.760877 u \times 931 MeV/u = 1639.375487 MeV

The binding energy per nucleon is $B E / A = \frac{B E}{A}$ .

B E / A = \frac{1639.375487 MeV}{209} \approx 7.8439 MeV/nucleon

Rounding to two decimal places, the binding energy per nucleon is $7.84 MeV/nucleon$ .

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Physics Question 44 – JEE-MAIN 2026

Clue 1 of 3: Nuclear Structure

🧠 Full Solution Path