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  • An electron is in an excited state in a hydrogen-like atom. It has a total energy of . The kinetic energy of the electron

An electron is in an excited state in a hydrogen-like atom. It has a total energy of -3.4 \mathrm{eV}. The kinetic energy of the electron is \mathrm{E} and its de Broglie wavelength is \lambda. Then,

Option: 1

\mathrm{E}=6.8 \mathrm{eV}, \lambda \sim 6.6 \times 10^{-10} \mathrm{~m}


Option: 2

\mathrm{E}=3.4 \mathrm{eV}, \lambda \sim 6.6 \times 10^{-10} \mathrm{~m}


Option: 3

\mathrm{E}=3.4 \mathrm{eV}, \lambda \sim 6.6 \times 10^{-11} \mathrm{~m}


Option: 4

\mathrm{E}=6.8 \mathrm{eV}, \lambda \sim 6.6 \times 10^{-11} \mathrm{~m}


Answers (1)

best_answer

The potential energy =-2 \times \text { kinetic energy }=-2

\therefore \text { Total energy }=-2 \mathrm{E}+\mathrm{E}=-\mathrm{E}=-3.4 \mathrm{eV}

\text { or } \mathrm{E}=3.4 \mathrm{eV}

Let p = momentum and m = mass of the electron.

\therefore \quad E=\frac{p^2}{2 m} \quad \text { or } p=\sqrt{2 m E}

\text { de-Broglie wavelength } \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=6.6 \times 10^{-10} \mathrm{~m}

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