Question

Determine the energy of 2.00 mol of photons for each of the following kinds of light. (Assume three significant figures.) You may want to reference (Pages 299 - 308) Section 7.2 while completing this problem. Part A infrared radiation (1460 nm )

Answer 1

The energy of 2.00 mol of photons of an infrared radiation 1460 nm is 164.08 kilo Joules.

Step-by-step explanation:

`E=(h* c)/(\lambda)`

where,

E = energy of a photon = ?

h = Planck's constant =
`6.63* 10^(-34)Js`

c = speed of light =
`3* 10^8m/s`

`\lambda` = wavelength =
`1460 nm^o=1460* 10^(-9)m`

Now put all the given values in the above formula, we get the energy of the photons.

`E=((6.63* 10^(-34)Js)* (3* 10^8m/s))/(1460* 10^(-9)m)`

`E=1.3623* 10^(-19)J`

1 mol =
`6.022* 10^(23)` particles

Then in 2 moles of photons have:

`2.00 mole= 2.00* 6.022* 10^(23) =1.2044* 10^(24)` photons

Total energy of the
`1.0244* 10^(24)` photons = E'

`E'=1.3623* 10^(-19) J* 1.0244* 10^(24)=164,075.41 J=164.08 kJ`

The energy of 2.00 mol of photons of an infrared radiation 1460 nm is 164.08 kilo Joules.