Energy of an electron in n^{th} orbit of hydr | vedclass

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(A) According to Bohr's postulates,the centripetal force required for an electron of mass $m$ and velocity $v$ to revolve in an orbit of radius $r$ is

Vedclass · Accepted Answer

$ - \frac{{2{\pi ^2}{k^2}m{e^4}}}{{{n^2}{h^2}}}$

Vedclass · Answer

(A) According to Bohr's postulates,the centripetal force required for an electron of mass $m$ and velocity $v$ to revolve in an orbit of radius $r$ is provided by the electrostatic force of attraction between the nucleus (charge $+e$) and the electron (charge $-e$): $\frac{mv^2}{r} = \frac{ke^2}{r^2} \implies mv^2 = \frac{ke^2}{r} \quad (1)$ From Bohr's quantization condition,the angular momentum is quantized: $mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr} \quad (2)$ Substituting $(2)$ into $(1)$: $m\left(\frac{nh}{2\pi mr}\right)^2 = \frac{ke^2}{r} \implies \frac{n^2h^2}{4\pi^2mr^2} = \frac{ke^2}{r} \implies r = \frac{n^2h^2}{4\pi^2kme^2}$ The kinetic energy $E_K$ is: $E_K = \frac{1}{2}mv^2 = \frac{1}{2}\left(\frac{ke^2}{r}\right) = \frac{ke^2}{2} \left(\frac{4\pi^2kme^2}{n^2h^2}\right) = \frac{2\pi^2k^2me^4}{n^2h^2}$ The potential energy $E_p$ is: $E_p = -\frac{ke^2}{r} = -ke^2 \left(\frac{4\pi^2kme^2}{n^2h^2}\right) = -\frac{4\pi^2k^2me^4}{n^2h^2}$ The total energy $E$ is the sum of kinetic and potential energy: $E = E_K + E_p = \frac{2\pi^2k^2me^4}{n^2h^2} - \frac{4\pi^2k^2me^4}{n^2h^2} = -\frac{2\pi^2k^2me^4}{n^2h^2}$

Energy of an electron in $n^{th}$ orbit of hydrogen atom is $\left( {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)$

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