Question

In Parts B and C you saw that, according to Bohr's postulate, the electron radius r and the electron velocity v only have certain allowable values. Plug the values obtained for these two quantities into the energy statement given below (E=) to arrive at a new statement for the allowed energy levels in the Bohr atom. Express answer in terms of e, m, n, h, \epsilon _{0} . Equations:
v=nh/2mr\pi

r=n^{2}h^{2}\epsilon _{0}/m\pi e^{2}

E=\frac{1}{2}mv^{2}-(e^{2}/4\pi r\epsilon _{0})

Answer 1

Answer:

$E=-\dfrac{ me^{4} }{8 n^2h^2\epsilon _{0}^2}$

Explanation:

Given:

$E=\dfrac{1}{2}mv^{2}-\dfrac{e^{2}}{4\pi r\epsilon _{0}}\\v=\dfrac{nh}{2mr\pi} \\r=\dfrac{n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}$

We want to substitute the values of r and v into the Energy Statement.

$E=\dfrac{1}{2}mv^{2}-\dfrac{e^{2}}{4\pi r\epsilon _{0}}\\v=\dfrac{nh}{2mr\pi} \\r=\dfrac{n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}$

First, we substitute r into v for simplicity.

$v=\dfrac{nh}{2mr\pi} ,r=\dfrac{n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}\\v=\dfrac{nh}{2m\left(\dfrac{n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}\right)\pi}\\=\dfrac{nh}{\left(\dfrac{2m\pi n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}\right)}\\=\dfrac{nhm\pi e^{2}}{2m\pi n^{2}h^{2}\epsilon _{0}}\\v=\dfrac{ e^{2}}{2 nh\epsilon _{0}}$

Then,

$r=\dfrac{n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}$

$v=\dfrac{ e^{2}}{2 nh\epsilon _{0}}$

$E=\dfrac{1}{2}m\left(\dfrac{ e^{2}}{2 nh\epsilon _{0}}\right)^{2}-\dfrac{e^{2}}{4\pi \left(\dfrac{n^{2}h^{2}\epsilon _{0}}{m\pi e^{2}}\right)\epsilon _{0}}$

$=\dfrac{m}{2}\left(\dfrac{ e^{2}}{2 nh\epsilon _{0}}\right)^{2}-\dfrac{e^{2}}{ \left(\dfrac{4\pin^{2}h^{2}\epsilon _{0}^2}{m\pi e^{2}}\right)}$

$=\dfrac{m}{2}\left(\dfrac{ e^{2}}{2 nh\epsilon _{0}}\right)^{2}-\dfrac{m\pi e^{2}e^{2}}{ 4\pin^{2}h^{2}\epsilon _{0}^2}$

$=\dfrac{m}{2}\cdot\dfrac{ e^{4}}{4 n^2h^2\epsilon _{0}^2}-\dfrac{m e^{4}}{ 4n^{2}h^{2}\epsilon _{0}^2}\\=\dfrac{ me^{4}}{8 n^2h^2\epsilon _{0}^2}-\dfrac{m e^{4}}{ 4n^{2}h^{2}\epsilon _{0}^2}$

$=\dfrac{ me^{4} -2me^{4}}{8 n^2h^2\epsilon _{0}^2}\\E=-\dfrac{ me^{4} }{8 n^2h^2\epsilon _{0}^2}$

The energy statement in terms of e,m, n,h and$\epsilon _{0}$ is:

$E=-\dfrac{ me^{4} }{8 n^2h^2\epsilon _{0}^2}$