Answer to Question #169412 in Atomic and Nuclear Physics for AFOLABI SODIQ

Question

Calculate the radius and energy of the electron in the nth orbital in the hydrogen atom

Answer 1

The radius and energy of the electron in the nth orbital in the hydrogen atom can be calculated in a semiclassical approximation using Bohr's postulates.

Start from the classical case. The electron is held in a circular orbit by electrostatic attraction. The centripetal force is equal to the Coulomb force:

${m_\mathrm{e} v^2\over r} = {Zk_\mathrm{e} e^2 \over r^2}$

where $m_e$ is the mass of the electron, $v$ is its speed, $r$ is the radius of the orbit, $Z$ is the atom's atomic number ($Z = 1$ for hydrogen), $k_e$ is the Coulomb constant, and $e$ is the charge of the electron.

Thus, the speed is:

$v = \sqrt{ k_\mathrm{e} e^2 \over m_\mathrm{e} r}$

and the energy is:

$E= -{1\over 2} m_\mathrm{e} v^2$

Now, let's use the Bohr's assumption - angular momentum is an integer multiple of $\hbar$ :

$m_evr = n\hbar$

Substituting the expression for the velocity into the last expression gives an equation for r in terms of n:

$m_{\text{e}}\sqrt{\dfrac{k_{\text{e}}e^2}{m_{\text{e}}r}}r=n\hbar\\ r_n = {n^2\hbar^2\over k_\mathrm{e} e^2 m_\mathrm{e}}$

The same procedure for the energy gives:

$E_n = -{k_\mathrm{e} e^2 \over 2r_n } = - { k_\mathrm{e}^2 e^4 m_\mathrm{e} \over 2\hbar^2 n^2}$

Answer. $E_n= - { k_\mathrm{e}^2 e^4 m_\mathrm{e} \over 2\hbar^2 n^2}, r_n = {n^2\hbar^2\over k_\mathrm{e} e^2 m_\mathrm{e}}$.