Question

pl answer with proper steps!!

Answer 1

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

We need Energy corresponding to photon of that wavelength:

\begin{lgathered}c = f\lambda \\ \\ \implies f = \frac{c}{\lambda}\end{lgathered}c=fλ⟹f=λc​​ 

Here,

c = Speed of light in vacuum = 3\times 10^8 \, \, m/s3×108m/s 
f = Frequency 
\lambdaλ = Wavelength = 230 nm = 230 \times 10^{-9} \, \, m230×10−9m 

Now, we have:

\begin{lgathered}E = hf \\ \\ \implies E = \frac{hv}{\lambda}\end{lgathered}E=hf⟹E=λhv​​ 
Here, 

h = Planck's Constant = 6.626\times 10^{-34} \, \, J s6.626×10−34Js

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 EIE​=NA​×E 


Where

N_ANA​ = Avogadro Number = 6.022 \times 10^{23}6.022×1023 


We can finally find the ionisation enthalpy:

\begin{lgathered}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}\end{lgathered}IE​=NA​×EandE=λhc​⟹IE​=NA​×λhc​⟹IE​=(6.022×1023)×230×10−96.626×10−34×3×108​J/mol⟹IE​≈520.46×103J/mol⟹IE​≈520.46kJ/mol​​


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

So, the Answer is Option a) 520