Q. If Avogadro number is $6 \times 10^{23}$, then number of protons, neutrons and electrons is 14 g of $\ _{6}C^{14}$ are respectively
A) $36 \times 10^{23},48 \times 10^{23},\ 36 \times 10^{23}$
B) $36 \times 10^{23},36 \times 10^{23},\ 36 \times 10^{23}$
C) $48 \times 10^{23},36 \times 10^{23},\ 48 \times 10^{23}$
D) $48 \times 10^{23},48 \times 10^{23},\ 36 \times 10^{23}$
Explanation
(a) Each atom of $\ _{6}C^{14}$ contains $6p,\ 6e$ and $8n$
$\therefore In\ 14\ g\ of\ \ _{6}C^{14}$
$p = 6 \times 6 \times 10^{23} = 36 \times 10^{23}$
$n = 8 \times 6 \times 10^{23} = 48 \times 10^{23}$
$e = p = 36 \times 10^{23}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.