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  • An electron is in an excited state in a hydrogen - like atom. It has a total energy of <img alt="\mathrm{-3 \cdot 4 \mathrm{ev}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B-3

An electron is in an excited state in a hydrogen - like atom. It has a total energy of \mathrm{-3 \cdot 4 \mathrm{ev}}. The kinetic energy of the electron is \mathrm{E} and its De -Broglie wavelength is \mathrm{\lambda-}

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^{-10} \mathrm{~m}}


Option: 4

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


Answers (1)

best_answer

We know that , 
\mathrm{\begin{aligned} P E & =-2(k \cdot \epsilon \cdot) \\ & =-2 E \end{aligned}}
\mathrm{\begin{aligned} \therefore \text { Total energy }=-2 E+E & =-E \\ & =-3.4 \mathrm{eV} \end{aligned}}
Now, 
\mathrm{\text { wave length } \begin{aligned} (\lambda) & =\frac{h}{\sqrt{2 \mathrm{mk}}} \\ \lambda & =6.6 \times 10^{-10} \mathrm{~m} \end{aligned}}

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avinash.dongre

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