Question

Calculate the energy in kilocalorie per mole of photons of an electromagnetic radiation having a wavelength of $7600\stackrel{\circ }{A}$.

• A. 33.56
• B. 37.56
• C. 47.35
• D. 42.35

Answer 1

The correct option is B 37.56

Given,
wavelength $\left(\lambda \right)=7600\stackrel{\circ }{A}=7600\times {10}^{-8}\text{cm}$ .
Speed of light $\left(c\right)=3\times {10}^{10}{\text{cm s}}^{-1}$
Therefore frequency, $v=\frac{c}{\lambda }=\frac{3\times {10}^{10}}{7600\times {10}^{-8}}=3.947\times {10}^{14}{\text{s}}^{-1}$

Energy of one photon $=hv$
$=6.62\times {10}^{-34}\times 3.947\times {10}^{14}\text{J}$
$=2.61\times {10}^{-19}\text{J}$
Energy of one mole of photons
$=2.61\times {10}^{-19}\times 6.023\times {10}^{23}$
$=15.72\times {10}^4\text{J}=157.2\text{kJ}$

Energy of one mole of photons in kilocalorie $=\frac{157.2}{4.185}=37.56\text{kcal per mol}$
$[\because 1\text{kcal}=4.185\text{kJ}]$