Question

Solve this:

The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the de-Broglie wavelength associated with it.
Foreign 2015

Answer 1

Dear Student,


$$\begin{gathered} KE=3.4eV=3.4*1.6*{10}^{-19}J \\ momentumP=\sqrt{2m*KE}=\sqrt{2*9.1*{10}^{-31}*3.4*1.6*{10}^{-19}} \\ P=\frac{h}{\lambda } \\ \sqrt{2*9.1*{10}^{-31}*3.4*1.6*{10}^{-19}}=\frac{h}{\lambda } \\ 9.95*{10}^{-25}=\frac{6.626*{10}^{-34}}{\lambda } \\ \lambda =0.666*{10}^{-9}m \\ Regards \end{gathered}$$