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(B) For an electron in a hydrogen-like atom,the total energy $(TE)$ is equal to the negative of the kinetic energy $(KE)$: $TE = -KE$.Given $TE = -3.4

Vedclass · Accepted Answer

$E = 3.4 \, eV, \lambda = 6.6 	imes 10^{-10} \, m$

Vedclass · Answer

(B) For an electron in a hydrogen-like atom,the total energy $(TE)$ is equal to the negative of the kinetic energy $(KE)$: $TE = -KE$.Given $TE = -3.4 \, eV$,we have $-3.4 \, eV = -KE$,which implies $KE = 3.4 \, eV$.Thus,$E = 3.4 \, eV$.The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p = \sqrt{2mE}$.Substituting the values: $h = 6.63 	imes 10^{-34} \, J \cdot s$,$m = 9.11 	imes 10^{-31} \, kg$,and $E = 3.4 	imes 1.6 	imes 10^{-19} \, J$.$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.11 	imes 10^{-31} 	imes 3.4 	imes 1.6 	imes 10^{-19}}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{9.92 	imes 10^{-49}}} \approx \frac{6.63 	imes 10^{-34}}{9.96 	imes 10^{-25}} \approx 0.665 	imes 10^{-9} \, m = 6.65 	imes 10^{-10} \, m$.Rounding to the given options,$\lambda \approx 6.6 	imes 10^{-10} \, m$.

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

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