The potential energy of an electron is defined as U = \(\frac 12\) mw2x2 and follows Bohr’s law. Radius of orbit as function of n depends on? (w is same constant)
To solve this problem, we need to determine how the radius of the electron's orbit, r, depends on the quantum number n when the potential energy is given by \( U = \frac{1}{2} mw^2 x^2 \).
First, we know from Bohr's model that the centripetal force required for circular motion is provided by the electric force between the electron and the nucleus. This gives the equation: \(F_c = \frac{m v^2}{r} = \frac{k e^2}{r^2}\), where \( k \) is Coulomb's constant, \( e \) is the charge of the electron, and \( v \) is its velocity.
The given potential energy suggests that the system follows a harmonic oscillator potential: \(U = \frac{1}{2} mw^2 r^2\). For a circular orbit, the potential energy is also related to the effective force trying to pull the electron towards the center.
In this setup, the frequency \( w \) relates to how tightly bound the electron is, akin to the angular frequency of a harmonic oscillator. However, we note the relationship of the Bohr model with the quantum number \( n \). According to Bohr's model, the energy levels are quantized and given by: \(E_n = - \frac{13.6 \, eV}{n^2}\).
Energy conservation, the expression for potential energy, and Bohr's quantization rule imply that: \(r_n \propto n^2\) (in the traditional Coulomb potential). However, here the harmonic oscillator-like potential modifies this.
The harmonic oscillator potential given effectively transforms the dependency as follows: \(r_n \propto \sqrt{n}\) in such a case.
Thus, the radius of the electron's orbit as a function of the principal quantum number \( n \) depends on the square root of \( n \): \(r_n \propto \sqrt{n}\).
Therefore, the correct answer is that the radius of the orbit as a function of \( n \) depends on \(\sqrt{n}\).
