Question 215: Atoms & Nuclei

Question

Find the Binding energy per neucleon for $${}_{50}^{120}Sn$$. Mass of proton m
p
= 1.00783 U, mass of neutron
m
n
= 1.00867 U and mass of tin nucleus m
Sn
= 119.902199 U. (take 1U = 931 MeV)

Accepted Answer

$$B.E. = \Delta m{c^2}$$
$$ = \Delta m 	imes 931$$
$$\Delta m = \left( {50 	imes 1.00783} ight) + \left( {70 	imes 1.00867} ight) - \left\{ {119.902199} ight\}$$
$$ = \left\{ {120.9984 - 119.902199} ight\}\,U$$
$$ = 1.1238\,U$$
$$BE = 1.1238\, 	imes 931 = 1046.2578\,MeV$$
BE per nucleon $$ \simeq $$ 1046/120 $$ \approx $$ 8.5 Mev

Answer

9.0 MeV

Answer

8.5 MeV

Answer

8.0 MeV

Answer

7.5 MeV

Atoms & Nuclei

Find the Binding energy per neucleon for ${}_{50}^{120}Sn$. Mass of proton m p = 1.00783 U, mass of neutron m n = 1.00867 U and mass of tin nucleus m Sn = 119.902199 U. (take 1U = 931 MeV)

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