Question

Electromagnetic radiation of wavelength 286 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy in kJ/mol.(Planck's constant (h) = $6.626\times {10}^{-34}Js)$

• A. $418.8kJmo{l}^{-1}$
• B. $209.4kJmo{l}^{-1}$
• C. $41.8kJmo{l}^{-1}$
• D. $814.2kJmo{l}^{-1}$

Answer 1

The correct option is A $418.8kJmo{l}^{-1}$

$\lambda =242nm=242\times {10}^{-9}mc=3\times {10}^{8}m{s}^{-1}E=hv=h\frac{c}{\lambda }=6.626\times {10}^{-34}\times \frac{3\times {10}^8}{242\times {10}^{-9}}=0.082\times {10}^{-17}J=0.082\times {10}^{-20}kJ\text{Energy per mole for ionisation}=0.082\times {10}^{-20}\times 6.022\times {10}^{23}=493.6kJmo{l}^{-1}$