The energy of one mole of photons of radiatio | vedclass

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(C) The energy of one mole of photons is given by the formula: $E = \frac{hcN_A}{\lambda}$ Substituting the given values: $E = \frac{6.63 	imes 10^{-

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$399$

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(C) The energy of one mole of photons is given by the formula: $E = \frac{hcN_A}{\lambda}$ Substituting the given values: $E = \frac{6.63 	imes 10^{-34} \ J \ s 	imes 3 	imes 10^8 \ m \ s^{-1} 	imes 6.02 	imes 10^{23} \ mol^{-1}}{300 	imes 10^{-9} \ m}$ $E = \frac{11.976 	imes 10^{-2} \ J \ mol^{-1}}{3 	imes 10^{-7}}$ $E = 3.9913 	imes 10^5 \ J \ mol^{-1}$ $E \approx 399 \ kJ \ mol^{-1}$

The energy of one mole of photons of radiation of wavelength $300 \ nm$ is ...... $kJ \ mol^{-1}$. (Given: $h = 6.63 \times 10^{-34} \ J \ s, N_A = 6.02 \times 10^{23} \ mol^{-1}, c = 3 \times 10^8 \ m \ s^{-1}$)

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