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

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$6.64 	imes 10^{-8} \ cm$

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$.
Given $E_n = -3.4 \ eV$,we have $-3.4 = -13.6 / n^2$,which gives $n^2 = 4$,so $n = 2$.
The radius of the $n^{th}$ orbit is $r_n = 0.529 	imes n^2 \ \mathring{A} = 0.529 	imes 4 	imes 10^{-8} \ cm = 2.116 	imes 10^{-8} \ cm$.
According to the Bohr postulate,the circumference of the orbit is equal to an integral multiple of the de-Broglie wavelength: $2\pi r = n\lambda$.
Thus,$\lambda = (2 	imes 3.1416 	imes 2.116 	imes 10^{-8}) / 2 \ cm$.
$\lambda = 3.1416 	imes 2.116 	imes 10^{-8} \ cm \approx 6.64 	imes 10^{-8} \ cm$.

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

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