Question

The radius of a strontium atom is 215 pm. How many strontium atoms would have to be laid side by side to span a distance of 2.10 mm?

Answer 1

Here we have to get the number of strontium atom needed to be laid side by side to span a distance of 2.10 mm.

9.03×10⁷ number of strontium atoms are needed to cover the distance of 2.10mm.

The radius of the strontium atom is 215 pm, thus the diameter of strontium atom is (2×215) = 430 pm.

Now we know 1 pm is equivalent to 10⁻⁹ mm. Thus 430 pm is equivalent to (430×10⁻⁹) = 4.3×10⁻⁷ mm.

Thus 1 strontium atom can occupy 4.3×10⁻⁷mm of space. Now the distance to be occupied by the lay down of the strontium atom side by side of the span is 2.10 mm.

Hence, $\frac{2.10}{4.3}$×10⁷ = 9.03×10⁷ number of atoms are needed.

Answer 2

Answer:

$4.883\times 10^6 atoms$ of strontium would be laid side by side to span a given distance.

Explanation:

Radius of strontium atom ,r= 215 pm = $2.15\times 10^{-10} m$

$1 pm = 10^{-12} m$

Diameter of the strontium atom = d = 2r:

d = $2\times 2.15\times 10^{-10} m=4.3\times 10^{-10} m$[/tex]

let the number of strontium atoms be x

Length of the chain of strontium atoms laid side by side = 2.10 mm

1 mm = 0.001 m

=2.10 mm = 0.001 × 2.10 m = $2.10\times 10^{-3} m$

$x\times d=2.10 mm$

$x\times 4.3\times 10^{-10} m=2.10\times 10^{-3} m$

$x=\frac{2.10\times 10^{-3} m}{4.3\times 10^{-10} m}=4.883\times 10^6 atoms$

$4.883\times 10^6 atoms$ of strontium would be laid side by side to span a given distance.