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An electron, in a hydrogen-like atom, is in an excited state. It has a total energy of $-3.4 \mathrm{eV}$. The kinetic energy and the de-Broglie wavelength of the electron are respectively
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$+3.4 \mathrm{eV}, 0.66 \times 10^{-9} \mathrm{~m}$
$E_{n}=-3.4 \mathrm{eV}$
The kinetic energyis equal to the magnitude of total energy in this case. $\therefore \mathrm{K.E}=+3.4 \mathrm{eV}$
The de Broglie wavelength of electron
$\begin{aligned} \lambda=\frac{h}{\sqrt{2 m K}} &=\frac{6.64 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}} \mathrm{eV} \\ &=0.66 \times 10^{-9} \mathrm{~m} \end{aligned}$
The kinetic energyis equal to the magnitude of total energy in this case. $\therefore \mathrm{K.E}=+3.4 \mathrm{eV}$
The de Broglie wavelength of electron
$\begin{aligned} \lambda=\frac{h}{\sqrt{2 m K}} &=\frac{6.64 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}} \mathrm{eV} \\ &=0.66 \times 10^{-9} \mathrm{~m} \end{aligned}$
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