Question

Calculate the energy (in kilocalorie per mol) of photons of an electromagnetic radiation of wavelength $7600\stackrel{o}{A}.$

• A. 37.5 kcal/mol
• B. 37.5 kcal/mol
• C. 375 kcal/mol
• D. 0.375 kcal/mol

Answer 1

The correct option is A 37.5 kcal/mol

$\lambda =7600\stackrel{o}{A}=7600\times {10}^{-8}cmc=3\times {10}^{10}cm/s\text{We know that},v=\frac{c}{\lambda }=\frac{3\times {10}^{10}}{7600\times {10}^{-8}}=3.947\times {10}^{14}{s}^{-1}\text{Energy of one photon}=hv=6.62\times {10}^{-27}\times 3.947\times {10}^{14}=2.6\times {10}^{-12}erg\text{Energy of one mole of photons}=2.61\times {10}^{-12}\times 6.022\times {10}^{23}=15.71\times {10}^{11}erg1erg=4.185\times {10}^{10}kcal\text{Energy of one mole of photons in kilocalorie}=\frac{15.71\times {10}^{11}}{4.185\times {10}^{10}}=37.538kcal/mol$