An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV . The k

Question

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

Professor PC Thomas Classes · Answer

(b): The potential energy $=-2 	imes$ kinetic energy $=-2 E$. \ ∴ Total energy $=-2 E+E=-E=-3.4 \mathrm{eV}$ or $E=3.4 \mathrm{eV}$. Let $p=$ momentum and $m=$ mass of the electron. $ 	herefore \quad E=\frac{p^2}{2 m} 	ext { or } p=\sqrt{2 m E} $ \ de-Broglie wavelength, $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}$ \ On substituting the values, we get $ \lambda=\frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 3.4 	imes 1.6 	imes 10^{-19}}}=6.6 	imes 10^{-10} \mathrm{~m} $