Question

An electromagnetic radiation of wavelength 242 nm is just sufficient to ionize a sodium atom. Calculate the ionization energy of sodium in kJ ${mol}^{-1}$.

• A. 294.5 KJ ${mol}^{-1}$
• B. 394.5 KJ ${mol}^{-1}$
• C. 494.5 KJ ${mol}^{-1}$
• D. 594.5 KJ ${mol}^{-1}$

Answer 1

The correct option is C

494.5 KJ ${mol}^{-1}$

We know that, c = $\nu \lambda$ $\Rightarrow$ $\nu =\frac{c}{\lambda }$

By Planck's theory:

E = $h\nu$ $\Rightarrow$ E = $h\frac{c}{\lambda }$

Now, energy associated with a photon of 242 nm:

$\frac{hc}{\lambda }$ = $\frac{6.626\times {10}^{-34}\times 3.0\times {10}^8}{242\times {10}^{-9}}$

= 8.21 $\times$ ${10}^{-19}$ J

Since one atom of Na requires = 8.21 $\times$ ${10}^{-19}$J for ionization, 6.023 $\times$ ${10}^{23}$ atoms of Na for ionization requires:

$\Rightarrow$8.21 $\times$ ${10}^{-19}$ $\times$ 6.023 $\times$ ${10}^{23}$ = 49.45 $\times$ ${10}^4$J

= 495.5 KJ ${mol}^{-1}$