Question

Photons of minimum energy 496 kJ/mol are needed to ionize sodium atoms. Calculate the lowest frequecy of light that will ionize a sodium atom.

• A. $1.24\times {10}^{14}{\text{s}}^{-1}$
• B. $1.24\times {10}^{15}{\text{s}}^{-1}$
• C. $2.48\times {10}^{15}{\text{s}}^{-1}$
• D. $2.48\times {10}^{14}{\text{s}}^{-1}$

Answer 1

The correct option is B $1.24\times {10}^{15}{\text{s}}^{-1}$

Given,
The minimum energy of photon required to ionize one mole of sodium atom = 496 kJ/mol

$\therefore$ The minimum energy of photon required to ionize one sodium atom $=\frac{496}{6.022\times {10}^{23}}\text{kJ}$

$E=h\nu$
Hence, $\text{Energy of photon}h\nu =\frac{496\text{kJ}}{6.022\times {10}^{23}}$

Therefore the required frequency $\nu =\frac{496\times {10}^3}{(6.022\times {10}^{23})(6.6\times {10}^{-34})}=1.24\times {10}^{15}Hz$