If Avogadro's number is $6\times {10}^{23}$, then the number of protons, neutrons and electrons in 14 g of ${}_6{C}^{14}$ are respectively:

The correct option is A

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

Since the number of protons, neutrons and electrons in an atom of ${}_6{C}^{14}$ are 6, 8 and 6 respectively. As 14 gm of ${}_6{C}^{14}$ contains $6\times {10}^{23}$ atoms, therefore the numbers of protons, neutrons and electrons in 14 gm of ${}_6{C}^{14}$ are

Protons = $6\times 6\times {10}^{23}=36\times {10}^{23}$

Neutrons = $8\times 6\times {10}^{23}=48\times {10}^{23}$

Electrons = $6\times 6\times {10}^{23}=36\times {10}^{23}$