Question

An Electron, in a Hydrogen Atom, in its ground state absorbs $1.5$ times as much energy as the minimum required for its escape $(\mathrm{I}.\mathrm{E}.,13.6\mathrm{ev})$ from the atom. Calculate the value of $\lambda$ for the emitted electron.

Answer 1

Step 1: Calculating incident energy:

• IE of hydrogen atom $=13.6\mathrm{eV}$
  
• Incident energy $$\begin{gathered} =1.5\times \mathrm{IE} \\ =1.5\times 13.6\mathrm{eV} \end{gathered}$$

Step 2: Calculating maximum kinetic energy:
$$\begin{gathered} \mathrm{Kinetic}\mathrm{energy}\mathrm{of}\mathrm{e}-\mathrm{ejected}\mathrm{out}=(1.5\times 13.6)-13.6 \\ =0.5\times 13.6 \\ {\mathrm{E}}_{\mathrm{k}}(\mathrm{Maximum}\mathrm{kinetic}\mathrm{energy})=0.5\times 13.6\mathrm{eV} \\ =6.8\times 1.6\times {10}^{-19}\mathrm{J}(1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{J}) \end{gathered}$$

Step 3: Wavelength calculation:
$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{Em}}}(\mathrm{m}-\mathrm{mass}\mathrm{of}\mathrm{electron}) \\ =\frac{6.62\mathrm{x}{10}^{-34}\mathrm{Js}}{{\left(2\mathrm{x}6.8\mathrm{x}9.1\mathrm{x}{10}^{-31}\mathrm{Kg}\mathrm{x}1.6\mathrm{x}{10}^{-19}\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ =\frac{6.62\mathrm{x}{10}^{-9}}{14.076} \\ \mathrm{\lambda }=4.71{\mathrm{A}}^{°} \end{gathered}$$