The light of wavelength 242 , nm is required | vedclass

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(B) Given: $\lambda = 242 \, nm = 242 	imes 10^{-9} \, m$. The ionization energy $(IE)$ is given by the formula: $IE = \frac{N_A hc}{\lambda}$. Subst

Vedclass · Accepted Answer

$494.5$

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(B) Given: $\lambda = 242 \, nm = 242 	imes 10^{-9} \, m$. The ionization energy $(IE)$ is given by the formula: $IE = \frac{N_A hc}{\lambda}$. Substituting the values: $N_A = 6.022 	imes 10^{23} \, mol^{-1}$,$h = 6.626 	imes 10^{-34} \, J \cdot s$,$c = 3 	imes 10^8 \, m \cdot s^{-1}$. $IE = \frac{6.022 	imes 10^{23} 	imes 6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{242 	imes 10^{-9}} \, J \cdot mol^{-1}$. $IE \approx \frac{11.97 	imes 10^{-2}}{242 	imes 10^{-9}} \approx 4.945 	imes 10^5 \, J \cdot mol^{-1}$. Converting to $kJ \cdot mol^{-1}$: $IE = 494.5 \, kJ \cdot mol^{-1}$.

The light of wavelength $242 \, nm$ is required to remove an electron from a $Na$ atom. What is the ionization energy of the sodium atom in $kJ \, mol^{-1}$?

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