Question

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.

Answer 1

\[\text { de Broglie wavelength }, \lambda = \frac{h}{\sqrt{2mE}}\]

\[\text { where }\]

\[\text { h = Plank's constant }= 6 . 626 \times {10}^{- 34} J s\]

\[\text { m = Mass of the electron } = 9 . 1 \times {10}^{- 31} kg\]

\[\text { E = Energy of the electron } = 3 . 4 eV = 3 . 4 \times 1 . 6 \times {10}^{- 19} J\]

\[ \Rightarrow \lambda = \frac{6 . 626 \times {10}^{- 34}}{\sqrt{2(9 . 1 \times {10}^{- 31} )(3 . 4 \times 1 . 6 \times {10}^{- 19} )}}\]

\[ \Rightarrow \lambda = 0 . 67 \times {10}^{- 9} m\]