JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 22)
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)
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
27
Explanation
$$\Delta$$m = (Zmp + (A $$-$$ Z)mn) $$-$$ MAl
= (13 $$\times$$ 1.00726 + 14 $$\times$$ 1.00866) $$-$$ 27.18846
= 27.21562 $$-$$ 27.18846
= 0.02716 u
E = 27.16 x $$\times$$ 10$$-$$3 J
= (13 $$\times$$ 1.00726 + 14 $$\times$$ 1.00866) $$-$$ 27.18846
= 27.21562 $$-$$ 27.18846
= 0.02716 u
E = 27.16 x $$\times$$ 10$$-$$3 J
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