Question

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

• A. $49.3KJmol{e}^{-1}$
• B. $0.493KJmol{e}^{-1}$
• C. $493.6kJmol{e}^{-1}$
• D. none of these

Answer 1

Answer: (C)

$493.6kJmol{e}^{-1}$

wavelength $\lambda =242nm=2.42\times {10}^{-7}m$

we know,

Energy $\left(E\right)=hc/\lambda$

here, h is Plank's constant

c is the speed of light

and $\lambda$ is the wavelength of photons

so,$E=\frac{6.625\times {10}^{-34}Js\times 3\times {10}^{8}m/s}{2.42\times {10}^{-}7m}$

$=0.0821\times {10}^{-17}$ J / atom

$0.0821\times {10}^{-17}$ J energy is sufficient for ionisation of one Na atom , so it is the ionisation energy of Na .

hence, $I.E=0.0821\times {10}^{-17}$ J/atom

$=0.0821\times {10}^{-17}\times 6.022\times {10}^{22}$ J/mol

$=493.5$ KJ/mol