Question

Ionization energy of gaseous $Na$ atoms is $495.5$ kJ mol${}^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is:

$[h=6.626\times {10}^{-34}Js,N_A=6.022\times {10}^{23}mo{l}^{-1}]$

• A. $7.50\times {10}^4{s}^{-1}$
• B. $4.76\times {10}^{14}{s}^{-1}$
• C. $3.15\times {10}^{15}{s}^{-1}$
• D. $1.24\times {10}^{15}{s}^{-1}$

Answer 1

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

As we know, $\frac{\Delta E}{N_A}=h\nu$
$\nu =\frac{\Delta E}{N_A\times h}$
$\nu =\frac{495.5\times {10}^3}{6.626\times {10}^{-34}\times 6.022\times {10}^{23}}=1.24\times {10}^{15}{s}^{-1}$