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)
Answer (integer)
27
Solution
$\Delta$m = (Zm<sub>p</sub> + (A $-$ Z)m<sub>n</sub>) $-$ M<sub>Al</sub><br><br>= (13 $\times$ 1.00726 + 14 $\times$ 1.00866) $-$ 27.18846<br><br>= 27.21562 $-$ 27.18846<br><br>= 0.02716 u<br><br>E = 27.16 x $\times$ 10<sup>$-$3</sup> J
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Nuclear Binding Energy
This question is part of PrepWiser's free JEE Main question bank. 184 more solved questions on Atoms and Nuclei are available — start with the harder ones if your accuracy is >70%.