If the number of neutrons in _{8}O^{16} is do | vedclass

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(C) The initial composition of $_{8}O^{16}$ is: Protons $(p)$ = $8$,Neutrons $(n)$ = $8$,Electrons $(e^-)$ = $8$. Mass is primarily determined by prot

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$50\, \%$ increase

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(C) The initial composition of $_{8}O^{16}$ is: Protons $(p)$ = $8$,Neutrons $(n)$ = $8$,Electrons $(e^-)$ = $8$. Mass is primarily determined by protons and neutrons. Initial mass number = $8 + 8 = 16$. New number of neutrons = $8 	imes 2 = 16$. New number of protons remains = $8$. New mass number = $16 + 8 = 24$. Change in mass = $24 - 16 = 8$. Percentage change = $(8 / 16) 	imes 100 = 50\, \% 	ext{ increase}$. Note: Electrons have negligible mass compared to protons and neutrons,so changing the number of electrons does not affect the mass calculation.

$(i) \ { }_{26} Fe ^{54}, { }_{26} Fe ^{56}, { }_{26} Fe ^{57}, { }_{26} Fe ^{58}$	$(a)$ Isotopes
$(ii) \ { }_1 H ^3, { }_2 He ^3$	$(b)$ Isotones
$(iii) \ { }_{32} Ge ^{76}, { }_{33} As ^{77}$	$(c)$ Isodiaphers
$(iv) \ { }_{92} U ^{235}, { }_{90} Th ^{231}$	$(d)$ Isobars
$(v) \ { }_1 H ^1, { }_1 D ^2, { }_1 T^3$

If the number of neutrons in $_{8}O^{16}$ is doubled and the number of electrons is halved,what is the percentage change in its mass?

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