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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$. Given $E_n = -3.4 	ext{ eV}$

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$6.6 	imes 10^{-10} 	ext{ m}$

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(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$. Given $E_n = -3.4 	ext{ eV}$,we have $-\frac{13.6}{n^2} = -3.4$,which gives $n^2 = 4$,so $n = 2$. According to Bohr's postulate,$mvr = \frac{nh}{2\pi}$,so the de-Broglie wavelength $\lambda = \frac{h}{mv} = \frac{2\pi r}{n}$. For hydrogen,the radius of the $n^{th}$ orbit is $r_n = 0.529 	imes n^2 	ext{ Å}$. For $n = 2$,$r_2 = 0.529 	imes 2^2 = 2.116 	ext{ Å}$. Substituting these values,$\lambda = \frac{2 	imes \pi 	imes 2.116}{2} = 6.64 	ext{ Å} = 6.64 	imes 10^{-10} 	ext{ m}$.

If the total energy of an electron in a hydrogen atom in an excited state is $-3.4 \text{ eV}$,then the de-Broglie wavelength of the electron is

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