Question

The electron energy in hydrogen atom is given by \(E_n = \frac {(–2.18×10^{–18})}{n^2} \ J\). 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 1

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

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