"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)"

Question

Accepted Answer

"$B.E. = \Delta m{c^2}$

$= \Delta m \times 931$

$$\Delta m = \left( {50 \times 1.00783} \right) + \left( {70 \times 1.00867} \right) - \left\{ {119.902199} \right\}$$

$= \left\{ {120.9984 - 119.902199} \right\}\,U$

$= 1.1238\,U$

$BE = 1.1238\, \times 931 = 1046.2578\,MeV$

BE per nucleon $\simeq$ 1046/120 $\approx$ 8.5 Mev"

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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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)

Solution

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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)