Q12-6P - Organic Chemistry (Mcmurry)

Question

Question: It’s useful to develop a feeling for the amounts of energy that correspond to different parts of the electromagnetic spectrum. Calculate the energies in kJ/mol of each of the following kinds of radiation: 

• (a) A gamma ray with $\lambda =5.0\times {10}^{-11}m$
• (b) An X-ray with $\lambda =3.0\times {10}^{-9}m$ 
• (c) Ultraviolet light with $v=6.0\times {10}^{15}Hz$
• (d) The visible light with $v=7.0\times {10}^{14}Hz$ .
• (e) Infrared radiation with $\lambda =2.0\times {10}^{-6}m$
• (f) Microwave radiation with $v=1.0\times {10}^{11}Hz$ .

Step-by-Step Solution

Answer

• (a) Energy in kJ/mol for gamma-ray is $2.4\times {10}^{6}kJ/mol$. 
• (b) Energy in kJ/mol for X-ray is $4\times {10}^{4}kJ/mol.$.
• (c) Energy in kJ/mol for ultraviolet light is $2.4\times {10}^{3}kJ/mol$. 
• (d) Energy in kJ/mol for visible light is $2.8\times {10}^{2}kJ/mol$ . 
• (e) Energy in kJ/mol for infrared radiation is $6.0kJ/mol$. 

          ( f) Energy in kJ/mol for microwave radiation is $4.0\times {10}^{-2}kJ/mol$.

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:

 

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

 

Where, 

 

E is the energy of the photon

v is the frequency of electromagnetic radiation and $v=\frac{c}{\lambda }$

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

 

$\lambda$ is the wavelength of electromagnetic radiation.

 

h is Planck's constant $=6.62\times {10}^{-34}Js$

 

 

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 energy will be.  

Step 3: Calculation of energy in part (a):

•  for , λ=5.0×10-11m            E=hcλ               =6.62×10-34×3×108×6.022×10235×10-11×1000              =2.4×106kJ/mol· To convert 1J to J/mol, multiply it by Avogadro’s number, i.e.,$6.022\times {10}^{23}$ .
  · $$\begin{gathered} \lambda =5.0\times {10}^{-11}m \\ E=\frac{hc}{\lambda } \\ =\frac{6.62\times {10}^{-34}\times 3\times {10}^8\times 6.022\times {10}^{23}}{5\times {10}^{-11}\times 1000} \\ =2.4\times {10}^{6}kJ/mol \end{gathered}$$

                     

Step 4: Calculation of energy in part (b):

• · To convert 1J to J/mol, multiply it by Avogadro's number, i.e., .

•   $$\begin{gathered} For, \lambda =3.0\times {10}^{-9}m \\ E=\frac{hc}{\lambda } \\ =\frac{6.62\times {10}^{-34}\times 3\times {10}^8\times 6.022\times {10}^{23}}{3\times {10}^{-9}\times 1000} \\ =4\times {10}^{4}kJ/mol \end{gathered}$$

                      

Step 5: Calculation of energy in part (c):

• · To convert 1J to J/mol, multiply it by Avogadro's number, i.e.,$6.022\times {10}^{23}$ .
  
  

            $$\begin{gathered} For, v=6.0\times {10}^{15}Hz \\ E=hv \\ =6.62\times {10}^{-34}\times 6.0\times {10}^{15}\times 6.022\times {10}^{23}\times {10}^{-3} \\ =2.4\times {10}^{3}kJ/mol \end{gathered}$$

       

 

Step 6: Calculation of energy in part (d):

• · To convert 1J to J/mol, multiply it by Avogadro's number, i.e., $6.022\times {10}^{23}$.

 

              $$\begin{gathered} For, v=7.0\times {10}^{14}Hz \\ E=hv \\ =6.62\times {10}^{-34}\times 7.0\times {10}^{14}\times 6.022\times {10}^{23}\times {10}^{-3} \\ =2.8\times {10}^{2}kJ/mol \end{gathered}$$

 

Step 7: Calculation of energy in part (e):

• · To convert 1J to J/mol, multiply it by Avogadro's number, i.e.,$6.022\times {10}^{23}$ .
• $$\begin{gathered} For, \lambda =2.0\times {10}^{-5}m \\ E=\frac{hc}{\lambda } \\ =\frac{6.62\times {10}^{-34}\times 3\times {10}^8\times 6.022\times {10}^{23}}{2\times {10}^{-5}\times 1000} \\ =6.0kJ/mol \end{gathered}$$· 

                     

Step 8: Calculation of energy in part (f):

• · To convert 1J to J/mol, multiply it by Avogadro's number, i.e.,$6.022\times {10}^{23}$ .
• $$\begin{gathered} For, v=1.0\times {10}^{11}Hz \\ E=hv \\ =6.62\times {10}^{-34}\times 1.0\times {10}^{11}\times 6.022\times {10}^{23}\times {10}^{-3} \\ =4.0\times {10}^{-2}kJ/mol \end{gathered}$$·