Question

The minimum energy needed to dissociate iodine molecules, I2, is 151 kJ/mol. What is the wavelength of photons (in nm) that supplies this energy, assuming each bond dissociated by absorbing one photon?

Answer 1

`792` nanometers (or scientifically `7.92*10^2*mum`.)

Explanation:

Equations this solution applied:

1. `N=n*N_A` where `N` is the quantity of `n` moles of particles and `N_A=6.02*10^23*"mol"^(-1)` is the Avagordoro's number

2. The Planck's Law `E=h*f` where `E` is the energy of a single photon of frequency `f` and `h` is the Planck's Constant, `h=6.63 × 10^(-34)* "m"^2* "kg" *"s"^(-1)= 6.63*10^(-34) color(blue)("J")*"s"` [1]

3. `lambda=v/f` where `lambda` is the wavelength of a wave or an electromagnetic (EM) radiation of frequency `f`.

From the question, breaking `N_A` (one mole of) iodine molecules consumes
`E("one mole")=E*N_A=151*color(red)("kJ")*"mol"^(-1)`
`=1.51*10^5*color(blue)("J") *"mol"^(-1)`
where `E` is the energy input required to break a single molecule.

Thus it takes
`E=(E("one mole"))/(N_A)=(1.51*10^5)/(6.02*10^23)=2.51* 10^(-19) * color(blue)("J")`
to break a single iodine molecule.

Apply the Planck's Law to find the maximum frequency of the EM radiation capable of breaking one such molecule:
`f=E/h= (2.51* 10^(-19) * color(blue)("J"))/(6.63*10^(-34) color(blue)("J")*"s")`
`=3.79*10^14 "s"^(-1)`

*Make sure that you get the unit that corresponds to the quantity after canceling out corresponding pairs. Here we are expecting `"s"^-1` for frequency, which does appear to be the case. *

Assuming `3.00*10^8 * color(magenta)("m")* "s"^-1=3.00\cdot 10^(17)* color(green)(mu "m")\cdot "s"^-1` to be the speed of light. A photon of such EM radiation is expected to have the wavelength
`lambda=v/f=(3.00*10^17* color(green)(mu"m")*"s"^(-1))/(3.79*10^14* "s"^(-1))=7.92*10^2* color(green)(mu "m)`

Sources:
1. Units ("dimensions") of the Planck's Constant: https://www.askiitians.com/forums/General-Physics/find-the-dimension-of-planck-constant-h-from-the-e_74309.htm