Q12-5P - Organic Chemistry (Mcmurry)

Question

Which has higher energy, infrared radiation with $\mathit{\lambda }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{m}$ or an X-ray with $\mathit{\lambda }\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{m}$? Radiation with $\mathit{\nu }\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathit{H}\mathit{z}$ or with $\mathit{\lambda }\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathit{m}$?

Step-by-Step Solution

Answer

X-ray has more energy than infrared radiation.

Radiation with $\lambda =9.0\times {10}^{-6} \mathrm{m}$ has more energy than $\nu =4.0\times {10}^9 \mathrm{Hz}$

Step 1: Quantum energy

A photon is a particle-like packet of electromagnetic energy. The formula to  calculate the energy (E) of a photon is equal to:

 

$\mathbf{E}\mathbf{=}\mathbf{h\nu }\mathbf{or}\frac{\mathbf{\text{hc}}}{\mathbf{\lambda }}$

 

Where, 

 

E is the energy of the photon

$\mathit{\nu }$ is the frequency of electromagnetic radiation and $\mathit{\nu }\mathbf{=}\frac{\mathbf{c}}{\mathbf{\lambda }}$

c is the speed of light in a vacuum $\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{m}\mathbf{/}\mathbf{s}$

 

$\mathit{\lambda }$ is the wavelength of electromagnetic radiation.

 

h is Planck's constant  $\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{62}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{34}}\mathbf{Js}$

Step 2: Factors affecting energy

From the formula, it can be seen energy depends on wavelength. Energy and wavelength are inversely proportional. So, the more the wavelength, the less will be energy.

Energy is directly proportional to the frequency of radiation. So, the more the frequency more will be energy.  

Step 3: Comparison

 

• From the given data, it can be seen that wavelength is more for infrared radiation than X-ray. So, X-ray has more energy than infrared radiation.
• For,  $\nu =4.0\times {10}^{9}Hz$

$$\begin{gathered} \mathrm{E}=\mathrm{h\nu } \\ =6.62\times {10}^{-34}\times 4.0\times {10}^9 \\ =2.65\times {10}^{-24}\mathrm{J} \end{gathered}$$

       

 

• For, $\mathrm{\lambda }=9.0\times {10}^{-6}\mathrm{m}$

                     

 $$\begin{gathered} \mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }} \\ =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{9\times {10}^{-6}} \\ =2.21\times {10}^{-20}\mathrm{J} \end{gathered}$$

•  From the above data, we can conclude radiation $\lambda =9.0\times {10}^{-6}$ with m has more energy than  $\mathrm{\nu }=4.0\times {10}^9\mathrm{Hz}$.