Question

If Avogadro number is $6\times 10^{23}$, then number of protons, neutrons and electrons is $14g$ 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. $48\times 10^{23},48\times 10^{23},36\times 10^{23}$

Answer 1

Answer: (A)

Each atom of ${}_6{C}^{14}$ contains $6p,6e$ and $8n$.
$\therefore \ln 14g$ of ${}_6{C}^{14}$
$p=6\times 6\times 10^{23}=36\times 10^{23}$
$n=8\times 6\times 10^{23}=48\times 10^{23}$
$e=p=36\times 10^{23}$