Question

Calculate the energy in kilocalorie per mole of photons of an electromagnetic radiation having a wavelength of $6600\stackrel{\circ }{A}$.

• A. 56.87 kcal/mol
• B. 43.32 kcal/mol
• C. 23.68 kcal/mol
• D. 65.46 kcal/mol

Answer 1

The correct option is B 43.32 kcal/mol

Given:
Wavelength $\left(\lambda \right)=6600\stackrel{\circ }{A}=6600\times {10}^{-10}\text{m}$
Speed of light $\left(c\right)=3\times {10}^8{\text{m s}}^{-1}$
Therefore frequency, $v=\frac{c}{\lambda }=\frac{3\times {10}^8}{6600\times {10}^{-10}}=4.54\times {10}^{14}{\text{s}}^{-1}$

Energy of one photon $=hv$
$=6.626\times {10}^{-34}\times 4.54\times {10}^{14}\text{J}$
$=3.01\times {10}^{-19}\text{J}$
Energy of one mole of photons
$=3.01\times {10}^{-19}\times 6.023\times {10}^{23}$
$=18.13\times {10}^4\text{J}=181.3\text{kJ}$

Energy of one mole of photons in kilocalorie $=\frac{181.3}{4.185}=43.32\text{kcal/mol}$
$[\because 1\text{kcal}=4.185\text{kJ}]$