Calculate binding energy per nucleon of ${}_{20}^{40}{Ca}$ nucleus.
(Given: mass of ${}_{20}^{40}{Ca} = 39.962589 u;$ mass of proton = 1.007825 u; mass of neutron = 1.00865 u; 1 atomic unit (1u) = 931 MeV)

Solution

Given, ${}_{20}^{40}{Ca}$ contains 20 protons and (40-20) = 20 neutrons.

Mass of 20 protons = 20 $\times$ 1.007825 = 20.1565 a.m.u
Mass of 20 neutrons = 20 $\times$ 1.008665 = 20.1733 a.m.u

Therefore, total mass of 40 nucleons = 40.3298 a.m.u

Given that mass of ${}_{20}^{40}{Ca}$ = 39.962589 a.m.u

Therefore, mass defect, $∆ m = 40.3298 - 39.962589 = 0.367211 a . m . u$

Total binding energy per nucleon is given by,

$\frac{0.367211 \times 931.5}{40} = 8.55 M e V$

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Wave Optics

Calculate binding energy per nucleon of ${}_{20}^{40}{Ca}$ nucleus.
(Given: mass of ${}_{20}^{40}{Ca} = 39.962589 u;$ mass of proton = 1.007825 u; mass of neutron = 1.00865 u; 1 atomic unit (1u) = 931 MeV)

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Wave Optics

Calculate binding energy per nucleon of Ca2040 nucleus.(Given: mass of Ca2040 = 39.962589 u; mass of proton = 1.007825 u; mass of neutron = 1.00865 u; 1 atomic unit (1u) = 931 MeV)

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Calculate binding energy per nucleon of ${}_{20}^{40}{Ca}$ nucleus.
(Given: mass of ${}_{20}^{40}{Ca} = 39.962589 u;$ mass of proton = 1.007825 u; mass of neutron = 1.00865 u; 1 atomic unit (1u) = 931 MeV)