Find the binding energy per nucleon for 50120 Sn. Mass of proton m p =1.00783 U , mass of neutron mn=1.00867 U and mass of tin nucleus m Sn =119.902199 U (take 1U = 931 MeV)

Question

Find the binding energy per nucleon for 50120 Sn. Mass of proton m p =1.00783 U , mass of neutron mn=1.00867 U and mass of tin nucleus m Sn =119.902199 U (take 1U = 931 MeV) (A) 8.5 MeV (B) 7.5 MeV (C) 8.0 MeV (D) 9.0 MeV

Tardigrade Admin · Accepted Answer

B.E. =[Δ m ] ⋅ c 2 M text expected = ZM p +( A - Z ) M n =50[1.00783]+70[1.00867] M text actual = 119.902199 B.E. =[50[1.00783]+70[1.00867]-119.902199] × 931 = 1020.56 ( BE / text nucleon ) =(1020.56/120) =8.5 MeV

Q. Find the binding energy per nucleon for $_{50}^{120} S n$ . Mass of proton $m_{p} = 1.00783 U,$ mass of neutron $m_{n} = 1.00867 U$ and mass of tin nucleus $m_{S n} = 119.902199 U$ (take $1 U = 931 M e V)$

$8.5 M e V$

$7.5 M e V$

$8.0 M e V$

$9.0 M e V$

B.E. $= [Δ m] \cdot c^{2}$
$M_{expected} = Z M_{p} + (A - Z) M_{n}$
$= 50 [1.00783] + 70 [1.00867]$
$M_{actual} = 119.902199$
$B . E . = [50 [1.00783] + 70 [1.00867] - 119.902199] \times 931$
$= 1020.56$
$\frac{BE}{nucleon} = \frac{1020.56}{120}$
$= 8.5 M e V$

Q. Find the binding energy per nucleon for 50120​Sn. Mass of proton mp​=1.00783U, mass of neutron mn​=1.00867U and mass of tin nucleus mSn​=119.902199U (take 1U=931MeV)

8.5MeV

7.5MeV

8.0MeV

9.0MeV