Calculate the binding energy per nucleon of 20 Ca 40 nucleus M ( 20 Ca 40)=39.962589 u , m n =1.008665 u m p =1.007825 u :

Question

Calculate the binding energy per nucleon of 20 Ca 40 nucleus M ( 20 Ca 40)=39.962589 u , m n =1.008665 u m p =1.007825 u : (A) 8.54 MeV / N (B) 9.54 MeV / N (C) 8.46 MeV / N (D) 10.56 MeV / N

Tardigrade Admin · Accepted Answer

Number of protons =20 Number of neutrons =40-20=20 Total mass of 20 protons and 20 Neutrons =20 m p +20 m n =20( m p + m m ) =20(1.007825+1.008665) =40.32984 Mass defect, Δ m =40.3298-39.962589 =0.367211 u Total B . E =0.367211 u =341.873441 mev B.E / N =(341.873441/41) =8.54 MeV / N

Q. Calculate the binding energy per nucleon of $_{20} C a^{40}$ nucleus $M (_{20} C a^{40}) = 39.962589 u, m_{n} = 1.008665 u$ $m_{p} = 1.007825 u :$

$8.54 M e V / N$

$9.54 M e V / N$

$8.46 M e V / N$

$10.56 M e V / N$

Q. Calculate the binding energy per nucleon of 20​Ca40 nucleus M(20​Ca40)=39.962589u,mn​=1.008665u mp​=1.007825u:

8.54MeV/N

9.54MeV/N

8.46MeV/N

10.56MeV/N

Number of protons =20 Number of neutrons =40−20=20 Total mass of 20 protons and 20 Neutrons =20mp​+20mn​=20(mp​+mm​) =20(1.007825+1.008665) =40.32984 Mass defect, Δm=40.3298−39.962589 =0.367211u Total B.E=0.367211u =341.873441mev B.E/N=41341.873441​ =8.54MeV/N

Q. Calculate the binding energy per nucleon of $_{20} C a^{40}$ nucleus $M (_{20} C a^{40}) = 39.962589 u, m_{n} = 1.008665 u$ $m_{p} = 1.007825 u :$

$8.54 M e V / N$

$9.54 M e V / N$

$8.46 M e V / N$

$10.56 M e V / N$