Electromagnetic radiation of wavelength 663 | vedclass

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Question

(D) The energy $E$ required to ionize one mole of atoms is given by the formula:$E = \frac{hcN_{A}}{\lambda 	imes 1000} 	ext{ (in } kJ \ mol^{-1}

Vedclass · Accepted Answer

$181$

Vedclass · Answer

(D) The energy $E$ required to ionize one mole of atoms is given by the formula:$E = \frac{hcN_{A}}{\lambda 	imes 1000} 	ext{ (in } kJ \ mol^{-1} 	ext{)}$Given:$h = 6.63 	imes 10^{-34} \ J \cdot s$$c = 3.00 	imes 10^{8} \ m \cdot s^{-1}$$N_{A} = 6.02 	imes 10^{23} \ mol^{-1}$$\lambda = 663 \ nm = 663 	imes 10^{-9} \ m$Substituting the values:$E = \frac{6.63 	imes 10^{-34} 	imes 3.00 	imes 10^{8} 	imes 6.02 	imes 10^{23}}{663 	imes 10^{-9} 	imes 1000}$$E = \frac{6.63 	imes 3.00 	imes 6.02 	imes 10^{-3}}{663 	imes 10^{-6}}$$E = \frac{119.799 	imes 10^{-3}}{663 	imes 10^{-6}} = \frac{119.799}{0.663} \approx 180.69 \ kJ \ mol^{-1}$Rounding off to the nearest integer,we get $181 \ kJ \ mol^{-1}$.

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