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, neutrons and electrons in 14 gm of ${}_6{C}^{14}$ are $6\times 6\times {10}^{23},=36\times {10}^{23},8\times 6\times {10}^{23}=48\times {10}^{23},6\times 6\times {10}^{23}=36\times {10}^{23}$.