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The energy of a hydrogen atom in the first excited state is -3.4 ev. Find :(a) the radius of this orbit. (Take Bohr radius \( =0.53 A \) )(b) the angular momentum of the electron in the orbit.(c) the kinetic and potential energy of the electron in the orbit. |
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Answer» Given E = -3.4 ev Then E = \(\frac{-13.6}{n^2}\) -3.4 = \(\frac{-13.6}{n^2}\) n2 = 4 n = 2 (a) r = \(\frac{e^2}{8\pi\varepsilon_0E_k}\) r = \(\frac{e^2}{4\pi\varepsilon_02E_k}\) r = \(\frac{9\times10^9\times(1.6\times10^{-19})^2}{2\times3.4}\) r = \(\frac{23.04\times10^{-29}}{6.8}\) r = 3.38 x 10-29m (b) Angular momentum L = \(\frac{nh}{2\pi}\) = \(\frac{2\times h}{2\pi}\) L = h/π 2.1 x 10-34 JS (c) For hydrogen atom Kinetic energy = -(total energy) Kinetic energy = 3.4 ev Potential energy = -2 x K.E. = - 2 x 3.4 P.E. = -6.8 ev |
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