Question.
Electromagnetic radiation of wavelength $242 \mathrm{~nm}$ is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $\mathrm{kJ} \mathrm{mol}^{-1}$.
Electromagnetic radiation of wavelength $242 \mathrm{~nm}$ is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in $\mathrm{kJ} \mathrm{mol}^{-1}$.
Solution:
Energy of sodium $(E)=\frac{N_{\Lambda} h c}{\lambda}$
$=\frac{\left(6.023 \times 10^{23} \mathrm{~mol}^{-1}\right)\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{242 \times 10^{-9} \mathrm{~m}}$
$=4.947 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$
$=494.7 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}$
$=494 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Energy of sodium $(E)=\frac{N_{\Lambda} h c}{\lambda}$
$=\frac{\left(6.023 \times 10^{23} \mathrm{~mol}^{-1}\right)\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{242 \times 10^{-9} \mathrm{~m}}$
$=4.947 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$
$=494.7 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}$
$=494 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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