JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 20)
Find the Binding energy per neucleon for $${}_{50}^{120}Sn$$. Mass of proton mp
= 1.00783 U, mass of neutron
mn
= 1.00867 U and mass of tin nucleus mSn = 119.902199 U. (take 1U = 931 MeV)
9.0 MeV
8.5 MeV
8.0 MeV
7.5 MeV
Explanation
$$B.E. = \Delta m{c^2}$$
$$ = \Delta m \times 931$$
$$\Delta m = \left( {50 \times 1.00783} \right) + \left( {70 \times 1.00867} \right) - \left\{ {119.902199} \right\}$$
$$ = \left\{ {120.9984 - 119.902199} \right\}\,U$$
$$ = 1.1238\,U$$
$$BE = 1.1238\, \times 931 = 1046.2578\,MeV$$
BE per nucleon $$ \simeq $$ 1046/120 $$ \approx $$ 8.5 Mev
$$ = \Delta m \times 931$$
$$\Delta m = \left( {50 \times 1.00783} \right) + \left( {70 \times 1.00867} \right) - \left\{ {119.902199} \right\}$$
$$ = \left\{ {120.9984 - 119.902199} \right\}\,U$$
$$ = 1.1238\,U$$
$$BE = 1.1238\, \times 931 = 1046.2578\,MeV$$
BE per nucleon $$ \simeq $$ 1046/120 $$ \approx $$ 8.5 Mev
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