Question

The radius of a uranium atom is 149 pm. How many uranium atoms would have to be laid side by side to span a distance of 4.96 mm?

Answer 1

Answer:

4960000000 pm

Explanation:

4.96*1000000000= 4960000000

Answer 2

Answer:

Uranium atoms required to be laid side by side to span a distance of 4.96 mm is $1.6644\times 10^7$ atoms.

Explanation:

The radius of a uranium atom ,r = 149 pm =$1.49\times 10^{-7} mm$

$1 pm=10^{-9} mm$

The diameter of a uranium atom ,d =

d = 2r =$2\times 1.49\times 10^{-7} mm=2.98\times 10^{-7} mm$

The let the uranium atom to be laid side by side to span a distance of 4.96 mm be x.

$x\times d=4.96 mm$

$x=\frac{4.96 mm}{d}=\frac{4.96 mm}{2.98\times 10^{-7} mm}=1.6644\times 10^7$ atoms.

Uranium atoms required to be laid side by side to span a distance of 4.96 mm is $1.6644\times 10^7$ atoms.