The potential energy of a proton and an elect | vedclass

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Question

(A) The potential energy $U$ is given by $U = eV = eV_0 \ln(r/r_0)$.The force $F$ is the negative gradient of the potential energy: $F = -\frac{dU}{dr

Vedclass · Accepted Answer

$r_n \propto n$

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(A) The potential energy $U$ is given by $U = eV = eV_0 \ln(r/r_0)$.The force $F$ is the negative gradient of the potential energy: $F = -\frac{dU}{dr} = -\frac{d}{dr} [eV_0 \ln(r/r_0)] = -\frac{eV_0}{r}$.The magnitude of the centripetal force is provided by this force: $\frac{mv^2}{r} = \frac{eV_0}{r}$.From this,we get $v^2 = \frac{eV_0}{m}$,which implies $v = \sqrt{\frac{eV_0}{m}}$. Note that $v$ is independent of $r$ and $n$.According to Bohr's quantization condition,the angular momentum is quantized: $mvr = \frac{nh}{2\pi}$.Substituting the expression for $v$: $mr \sqrt{\frac{eV_0}{m}} = \frac{nh}{2\pi}$.Rearranging for $r$: $r = \frac{nh}{2\pi} \sqrt{\frac{1}{meV_0}}$.Since $h, m, e, V_0$ are constants,we have $r_n \propto n$.

The potential energy of a proton and an electron in a hydrogen atom is given by $V = V_0 \ln(r/r_0)$,where $r_0$ is a constant. Assuming the Bohr model is applicable to this system,find the relationship between the radius $r_n$ and the principal quantum number $n$.

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