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

4960000000 pm

Step-by-step explanation:

4.96*1000000000= 4960000000

Answer 2

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

Step-by-step explanation:

The radius of a uranium atom ,r = 149 pm =
`1.49* 10^(-7) mm`

`1 pm=10^(-9) mm`

The diameter of a uranium atom ,d =

d = 2r =
`2* 1.49* 10^(-7) mm=2.98* 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* d=4.96 mm`

`x=(4.96 mm)/(d)=(4.96 mm)/(2.98* 10^(-7) mm)=1.6644* 10^7` atoms.

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