JEE Mains · Physics · STD 12 - 13. Nuclei
From the given data, the amount of energy required to break the nucleus of aluminium \({ }_{13}^{27} {Al}\) is \(....\,x \times 10^{-3} {J}\) Mass of neutron \(=1.00866 \,{u}\) Mass of proton \(=1.00726 \,{u}\) Mass of aluminium nucleus \(=27.18846\, {u}\) (Assume \(1\,u\) corresponds to \(x\,J\) of energy) (Round off to the nearest integer)
- A \(25\)
- B \(26\)
- C \(27\)
- D \(31\)
Answer & Solution
Correct Answer
(C) \(27\)
Step-by-step Solution
Detailed explanation
\(\Delta m =\left( Zm m _{ P }+( A - Z ) m _{ n }\right)- M _{ Al}\) \(=(13 \times 1.00726+14 \times 1.00866)-27.18846\) \(=27.21562-27.18846\) \(=0.02716 u\) \(E =27.16 \times 10^{-3} J\)
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