The ionisation energy of a certain element is | vedclass

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(B) The energy difference between the ground state $(GS)$ and the first excited state $(1^{st} ES)$ is given by the difference in their respective ion

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$UV$

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(B) The energy difference between the ground state $(GS)$ and the first excited state $(1^{st} ES)$ is given by the difference in their respective ionisation energies: $\Delta E = IE_{GS} - IE_{1^{st} ES} = 1540 \ kJ \ mol^{-1} - 300 \ kJ \ mol^{-1} = 1240 \ kJ \ mol^{-1}$.To convert this energy into electron volts $(eV)$,we use the conversion factor $1 \ eV \approx 96.5 \ kJ \ mol^{-1}$:$\Delta E = \frac{1240}{96.5} \approx 12.85 \ eV$.Using the relation $\lambda (nm) = \frac{1240}{\Delta E (eV)}$:$\lambda = \frac{1240}{12.85} \approx 96.5 \ nm$.Since the wavelength is approximately $96.5 \ nm$,which falls in the range of $10 \ nm$ to $400 \ nm$,the emitted light lies in the $UV$ region.

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