Question

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

• A. $36\times {10}^{23},48\times {10}^{23},36\times {10}^{23}$
  
• B. $36\times {10}^{23},36\times {10}^{23},36\times {10}^{21}$
• C. $48\times {10}^{23},36\times {10}^{23},48\times {10}^{21}$
• D. $48\times {10}^{23},48\times {10}^{23},36\times {10}^{21}$

Answer 1

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 in 14 gm of ${}_6{C}^{14}$ are $6\times 6\times {10}^{23}=36\times {10}^{23}$.

The numbers of neutrons in 14 gm of ${}_6{C}^{14}$ are $8\times 6\times {10}^{23}=48\times {10}^{23}$.

The numbers of electrons in 14 gm of ${}_6{C}^{14}$ are $6\times 6\times {10}^{23}=36\times {10}^{23}$.