Question

Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium atoms side by side in contact with one another. How many atoms would it take to make the distance 6.20 cm from end to end?

Answer 1

Answer:

$4.4285\times 10^8$ many atoms would it take to make the distance 6.20 cm from end to end.

Explanation:

Diameter of the helium atom =  $d=1.40\times 10^2 pm$

Let the number of atoms required to make the distance 6.20 cm be n.

$d\times n= 6.20 cm =6.20\times 10^{10} pm$

$1.40\times 10^2 pm\times n=6.20\times 10^{10} pm$

$n=\frac{6.20\times 10^{10} pm}{1.40\times 10^2 }=4.4285\times 10^8$

$4.4285\times 10^8$ many atoms would it take to make the distance 6.20 cm from end to end.