Question

Compare the energy (in joules) carried by an X-ray photon (wavelength = 0.20 nm) with that carried by an AM radio wave photon ( $\mathbf{\lambda }$= 200 m). Calculate the energy of 1.00 mol of each type of photon. What effect do you expect each type of radiation to have for inducing chemical reactions in the substances through which it passes?

Short answer

The energy of of X-ray photons is $5.98\times {10}^8{\text{Jmol}}^{-1}$, which is higher than the energy of 1 mole of AM radio wave photons, $5.98\times {10}^{-4}{\text{Jmol}}^{-1}$. Thus, X-ray photons can induce chemical reactions than AM radio wave photons as it has higher energy.

Step-by-step solution

Explaining the relation between the energy and the wavelength of a photon

The energy of each photon is inversely proportional to its wavelength. The shorter the wavelength, the higher the energy of the photon.

The mathematical relation between energy and wavelength is given as:

$E=\frac{hc}{\lambda }$

Where,

$$\begin{gathered} E=Energyofphoton \\ \text{h}=\text{Planck's}\text{constant}\text{=}\text{6}.\text{625}\times {\text{10}}^{\text{- 34}}\text{Js} \\ \text{c}=S\text{peed of light}=3\times {10}^8{\text{ms}}^{-1} \\ \mathbf{\lambda }=W\text{avelength} \end{gathered}$$

Calculating the energy of X-ray photon

The wavelength of X-ray photon $=0.20\text{nm}$or$2\times {10}^{-10}\text{m}$ .

Substituting values in the formula:

role="math" localid="1663404388398" $$\begin{gathered} E_{AM}=\frac{hc}{{\lambda }_{MM}} \\ {E}_{AM}=\frac{6.625\times {10}^{-34}\text{Js}\times 3\times {10}^8{\text{ms}}^{-1}}{200\text{m}} \\ {E}_{X-\text{ray}}=9.94\times {10}^{-28}\text{J} \end{gathered}$$

The energy of one X-ray photon $=9.94\times {10}^{-16}\text{J}$

1 mole of photon $=602x{10}^3$number of photons

Hence, energy of of X-ray photon is:

$$\begin{gathered} E=\frac{9.94\times {10}^{-16}\text{J}}{1\text{photon}}\times \frac{6.02\times {10}^{23}\text{photons}}{1\text{mol}} \\ {E}_{1\text{mol}}=5.98\times {10}^8{\text{Jmol}}^{-1} \end{gathered}$$

Calculating the energy of AM radio wave photon 

Wavelength of AM radio wave photon = 200m

Substituting value in formula as:

$$\begin{gathered} E_{AM}=\frac{hc}{{\lambda }_{AM}} \\ {E}_{AM}=\frac{6.625\times {10}^{-34}\text{Js}\times 3\times {10}^8{\text{ms}}^{-1}}{200\text{m}} \\ {E}_{AM}=9.94\times {10}^{-28}\text{J} \end{gathered}$$

Energy of one AM photon $=9.94\times {10}^{-28}\text{J}$.

1 mole of photon $=6.02\times {10}^{23}$number of photons

Hence, energy ofof AM photon is:

$$\begin{gathered} E=\frac{9.94\times {10}^{-28}\text{J}}{1\text{photon}}\times \frac{6.02\times {10}^{23}\text{photons}}{1\text{mol}} \\ {E}_{1\text{mol}}=5.98\times {10}^{-4}{\text{Jmol}}^{-1} \end{gathered}$$

Therefore, the energy of X-ray photons is, $5.98\times {10}^8{\text{Jmol}}^{-1}$, which is higher than the energy $1mole$of AM radio wave photons,$5.98\times {10}^{-4}{\text{Jmol}}^{-1}$ .

Thus, X-ray photons can induce more chemical reactions than AM radio wave photons as they have higher energy.