The potential energy of an electron in an orb | vedclass

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(B) The potential energy $U$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $U = -27.2 / n^2 	ext{ eV}$.Given $U = -6.8 	ext{ e

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$4 \pi r_0$

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(B) The potential energy $U$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $U = -27.2 / n^2 	ext{ eV}$.Given $U = -6.8 	ext{ eV}$,we have $-27.2 / n^2 = -6.8$,which implies $n^2 = 27.2 / 6.8 = 4$,so $n = 2$.The radius of the $n^{th}$ orbit is given by $r_n = n^2 r_0$. For $n = 2$,$r_2 = 2^2 r_0 = 4 r_0$.According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.Substituting the values,$2 \pi (4 r_0) = 2 \lambda$.Solving for $\lambda$,we get $\lambda = (8 \pi r_0) / 2 = 4 \pi r_0$.

The potential energy of an electron in an orbit of a hydrogen atom is $-6.8 \text{ eV}$. The de Broglie wavelength of the electron in this orbit is (where $r_0$ is the Bohr radius).

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