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Q.

Calculate the binding energy per nucleon of 2040Ca  .  Given that mass of 2040Ca  nucleus = 39.962589 u. mass of a proton = 1.007825 u ; mass of neutron = 1.008665 u and 1 u is equivalent to 931 Mev.

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a

4.55 Mev

b

8.55 Mev

c

6.55 Mev

d

7.55 Mev

answer is B.

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Detailed Solution

2040Ca   ,    Z=20 ;  A−Z=20 Δm=Zmp+(A−Z)mn−M          =20×1.007825+20×1.008665−39.962589          =0.367211amu BE=Δm×931.5Mev=342.06Mev BE/Nucleon=342.0640=8.55Mev
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