Question

What is the result in kilojoules of 5.0x10 atoms emitting energy for 2.00 minutes if Plank's 1 x atom (Hint: same hint as #13) ball part 3x10" 18 constant is 6.626 X 10-14

Answer 1

`3.973* 10^(-14) kJ` is the result in kiloJoules.

Step-by-step explanation:

Value of given Planck constant = h=
`6.626* 10^(-34) J s/ atom`

Energy emitted by 1 atom in a 1 sec :
`=6.626* 10^(-34) J`

2.00 min =
`2.00* 60 seconds = 120.00 seconds`

Energy emitted by 1 atom in 120 seconds = E'

`E'=6.626* 10^(-34)J * 120`

Let the energy emitted by
`5.0* 10^(20) atoms` in 2.00 minutes be E.

`E=E'* \text{Number of atoms}`

`E=6.626* 10^(-34) J* 120 * 5.0* 10^(20)`

`E=3.973* 10^(-11) J=3.973* 10^(-14) kJ`

(1 J = 0.001 kJ)

`3.973* 10^(-14) kJ` is the result in kiloJoules.