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

`E=-( me^(4) )/(8 n^2h^2\epsilon _(0)^2)`

Step-by-step explanation:

Given:

`E=(1)/(2)mv^(2)-(e^(2))/(4\pi r\epsilon _(0))\\v=(nh)/(2mr\pi) \\r=(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=(1)/(2)mv^(2)-(e^(2))/(4\pi r\epsilon _(0))\\v=(nh)/(2mr\pi) \\r=(n^(2)h^(2)\epsilon _(0))/(m\pi e^(2))`

First, we substitute r into v for simplicity.

`v=(nh)/(2mr\pi) ,r=(n^(2)h^(2)\epsilon _(0))/(m\pi e^(2))\\v=(nh)/(2m\left((n^(2)h^(2)\epsilon _(0))/(m\pi e^(2))\right)\pi)\\=(nh)/(\left((2m\pi n^(2)h^(2)\epsilon _(0))/(m\pi e^(2))\right))\\=(nhm\pi e^(2))/(2m\pi n^(2)h^(2)\epsilon _(0))\\v=( e^(2))/(2 nh\epsilon _(0))`

Then,

`r=(n^(2)h^(2)\epsilon _(0))/(m\pi e^(2))`

`v=( e^(2))/(2 nh\epsilon _(0))`

`E=(1)/(2)m\left(( e^(2))/(2 nh\epsilon _(0))\right)^(2)-(e^(2))/(4\pi \left((n^(2)h^(2)\epsilon _(0))/(m\pi e^(2))\right)\epsilon _(0))`

`=(m)/(2)\left(( e^(2))/(2 nh\epsilon _(0))\right)^(2)-(e^(2))/( \left((4\pin^(2)h^(2)\epsilon _(0)^2)/(m\pi e^(2))\right))`

`=(m)/(2)\left(( e^(2))/(2 nh\epsilon _(0))\right)^(2)-(m\pi e^(2)e^(2))/( 4\pin^(2)h^(2)\epsilon _(0)^2)`

`=(m)/(2)\cdot( e^(4))/(4 n^2h^2\epsilon _(0)^2)-(m e^(4))/( 4n^(2)h^(2)\epsilon _(0)^2)\\=( me^(4))/(8 n^2h^2\epsilon _(0)^2)-(m e^(4))/( 4n^(2)h^(2)\epsilon _(0)^2)`

`=( me^(4) -2me^(4))/(8 n^2h^2\epsilon _(0)^2)\\E=-( me^(4) )/(8 n^2h^2\epsilon _(0)^2)`

The energy statement in terms of e,m, n,h and
`\epsilon _(0)` is:

`E=-( me^(4) )/(8 n^2h^2\epsilon _(0)^2)`