Question

Question #15b3b

Answer 1

For the x-rays:

`E= 8.9xx10^5 (kJ)/"mol"`

For the `gamma`-rays

`E= 4.4xx10^9 (kJ)/"mol"`

Explanation:

To calculate the energy of these photons, we need the Planck relation written in terms of wavelength (`lambda`)

`E=(hc)/λ`

Since the units on ionization energy are kJ per mol, we must convert Planck's constant into a "kJ s per mol" value.

`h=(6.63×10^(−34)Js)⋅(6.02×10^23mol)÷(1000J/(kJ))=3.99×10^(−13) (kJs)/"mol"`

Using `C=3.00xx10^8 m/s`, the relation becomes

`E=(3.99×10^(−13) (kJs)/"mol"xx3.0xx10^8m/s)/λ=(1.20xx10^(-4))/lambda ((kJm)/"mol")`

Now, just insert the wavelength:

For the x-rays:

`E=(1.20xx10^(-4))/(0.135xx10^(-9)) = 8.9xx10^5 (kJ)/"mol"`

For the `gamma`-rays

`E=(1.20xx10^(-4))/(2.74xx10^(-14)) = 4.4xx10^9 (kJ)/"mol"`

In both cases, this energy is orders of ten greater than the typical bond energy of molecules in organic compounds (about 100 kJ per mol). This explains why these forms of radiation are so dangerous to living organisms.