Question

Radiation of wavelength 242 nm is sufficient to ionize the sodium atom. Calculate the ionization energy in $kJmo{l}^{-1}.(n=6.625\times {10}^{-34}J-s)$

• A. $493.6kJ$
• B. $6.625\times {10}^{3}J$
• C. $242kJ$
• D. $1324kJ$

Answer 1

The correct option is A $493.6kJ$

Ionisation energy is given as: $E=h\frac{c}{\lambda }$ ;

where h = $6.6\times {10}^{-34}J{s}^{-1}$, $\lambda =242\times {10}^{-9}m$, c = $3\times {10}^{8}m{s}^{-1}$

$\therefore E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{242\times {10}^{-9}}=4.936\times {10}^{5}J=493.6kJ$