Question

If electromagnetic radiation of wavelength 230 nm is just sufficient to ionize the lithium atom, then ionization energy of lithium atom in KJmol^-1 will be approximately
a)520
b)630
c)1040
d)2080

Answer 1

We are given that radiation of wavelength 230 nm ionizes one Lithium Atom.

We need Energy corresponding to photon of that wavelength:

$c = f\lambda \\ \\ \implies f = \frac{c}{\lambda}$

Here,

c = Speed of light in vacuum = $3\times 10^8 \, \, m/s$
f = Frequency
$\lambda$ = Wavelength = 230 nm = $230 \times 10^{-9} \, \, m$

Now, we have:

$E = hf \\ \\ \implies E = \frac{hv}{\lambda}$
Here, 

h = Planck's Constant = $6.626\times 10^{-34} \, \, J s$

This formula gives us the Energy required to ionize one Lithium Atom.

We need Ionization Enthalpy, which is measured in kJ / mol.


So, we need the energy to ionize one mole of Lithium Atoms.

Since One Atom takes energy E, one mole of Lithium Atoms would have the Ionization Energy of:

$I_E = N_A \times E$


Where

$N_A$ = Avogadro Number = $6.022 \times 10^{23}$


We can finally find the ionisation enthalpy:

$I_E = N_A \times E \quad and \quad E = \frac{hc}{\lambda} \\ \\ \\ \implies I_E = N_A \times \frac{hc}{\lambda} \\ \\ \\ \implies I_E = (6.022 \times 10^{23}) \times \frac{6.626\times 10^{-34} \times 3 \times 10^8}{230 \times 10^{-9}} \, \, \, J / mol \\ \\ \\ \implies I_E \approx 520.46 \times 10^3 \, \, \, J/mol \\ \\ \\ \implies \boxed{I_E \approx 520.46 \, \, \, kJ/mol}$


Thus, The Ionization Enthalpy/Energy of Lithium is 520.46 kJ/mol

So, the Answer is Option a) 520

Answer 2

Answer:

If electromagnetic radiation of wavelength 230 nm is just sufficient to ionize the lithium atom, then ionization energy of lithium atom in KJmol^-1 will be approximately. a)520.