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Chemistry Question on Quantum Mechanical Model of Atom

The electron energy in hydrogen atom is given by En=(–2.18×10–18)n2 JE_n = \frac {(–2.18×10^{–18})}{n^2} \ J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Answer

Given, En=−2.18×10−18n2 JE_n = \frac {-2.18×10^{-18}}{n^2}\ J
Energy required for ionization from n = 2 is given by ,
ΔE=E∞−E2ΔE = E_∞ - E_2
ΔE=[(−2.18×10−18(∞)2)−(−2.18×10−18(2)2)]JΔE= [\frac {(-2.18×10^{-18}}{(∞)^2}) - \frac {(- 2.18×10^{-18}}{(2)^2})] J

ΔE=[−2.18×10−184−0]JΔE= [ \frac {- 2.18×10^{-18}}{4} - 0 ] J
ΔE=0.545×10−18JΔE= 0.545×10^{-18} J
ΔE=5.45×10−19JΔE = 5.45×10^{-19} J
λ=hcΔEλ = \frac {hc}{ΔE}
Here, λ is the longest wavelength causing the transition.
λ=(6.626×10−34)(3×108)5.45×10−19λ =\frac {(6.626×10^{-34}) (3×10^8)}{5.45 × 10^{-19}}
λ=3.647×10−7mλ= 3.647×10^{-7} m
λ=3647×10−10 mλ = 3647×10^{-10}\ m
λ=3647 A˚λ = 3647\ Å