JEE MAIN - Chemistry (2024 - 31st January Morning Shift - No. 23)
The ionization energy of sodium in $$\mathrm{~kJ} \mathrm{~mol}^{-1}$$, if electromagnetic radiation of wavelength $$242 \mathrm{~nm}$$ is just sufficient to ionize sodium atom is _______.
Answer
494
Explanation
$$\begin{aligned}
& \mathrm{E}=\frac{1240}{\lambda(\mathrm{nm})} \mathrm{eV} \\
& =\frac{1240}{242} \mathrm{eV} \\
& =5.12 \mathrm{eV} \\
& =5.12 \times 1.6 \times 10^{-19} \\
& =8.198 \times 10^{-19} \mathrm{~J} / \text { atom } \\
& =494 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$$
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