Question 359: Atoms & Nuclei

Question

From the given data, the amount of energy required to break the nucleus of aluminium $$_{13}^{27}$$Al is __________ x $$	imes$$ 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)

Accepted Answer

$$\Delta$$m = (Zm
p
+ (A $$-$$ Z)m
n
) $$-$$ M
Al
= (13 $$	imes$$ 1.00726 + 14 $$	imes$$ 1.00866) $$-$$ 27.18846
= 27.21562 $$-$$ 27.18846
= 0.02716 u
E = 27.16 x $$	imes$$ 10
$$-$$3
J

Atoms & Nuclei

Integer / Numeric Answer Type