Question

If Avogadro number is $6\times {10}^{23}$, then number of protons, neutrons and electrons in 14 gms 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)

$36\times {10}^{23}$, $48\times {10}^{23}$, $36\times {10}^{23}$

Each atom of ${{}_{6}C}^{14}$ contains $6p,6eand8n$
$\therefore$ In 14 gram of ${{}_{6}C}^{14}$
$p=$ $6\times 6\times {10}^{23}$=$36\times {10}^{23}$
$n=$ $8\times 6\times {10}^{23}$=$48\times {10}^{23}$
$e=p=$ $36\times {10}^{23}$