lecture 12

Lecture 12
Bohr obtained values of the energies of various states or levels of the hydrogen atom by assuming that an electron with velocity v rotates in a circular orbit of radius r around the nucleus with an angular momentum of
m_e vr=nh/2?=n?, n=1,2,3,….. …………. (2-2)
This relationship expresses Bohr s second postulate in mathematical form. The electron mass is m_e, and the integer n is called the principal quantum number.
By using Newton s second law of motion, and equating the electrical force to the radial acceleration associated with the angular rotation of the electron around the much heavier nucleus (a proton):
(e.e)/(4??_o r^2 )=(m_e v^2)/r ……………… (2-3)
where e is the electron charge.
Using (2-2) and (2-3) to eliminate v leads to the formula for quantized orbits of the electron (Bohr orbits) described by the following radii:
r=(?_o h^2)/(?m_e e^2 ) n^2=a_H n^2 …………. (2-4)
where a_H is the radius of the first Bohr orbit.
Find the value of Bohr’s orbit when (n = 1)
The total energy for a specific orbit can be obtained by summing the kinetic energy and the potential energy of the electron with respect to the nucleus:
E=1/2 m_e v^2-e^2/(4??_o r) …………. (2-5)
where the first term is the kinetic energy of angular motion and the second term is the potential electrical energy of attraction, using the value of r given in (2-4) as the separation distance of the positive and negative charges. Eliminating v by means of (3-3) gives the following value for the energy:
E=-e^2/(8??_o r) ……………. (2-6)
Substituting the expression (3-4) for r into (3-6), we find that the electron can have only certain discrete negative values of energy E_n associated with the various values of n, where n is a positive integer such that
E_n=-(m_e e^4)/(8??_o^2 h^2 )?1/n^2 =-E_o/n^2 …………... (2-7)
Find the value of E_o
For a slightly more accurate value of the energy we must use the reduced mass ? of the combined electron and proton system,
?=m_e/(1+?m_e/m?_p )=(m_e m_p)/(m_e+m_p ) …………….. (2-8)
instead of the mass of the electron as in (2-7). The reduced mass takes into account the finite size of the nucleus and the fact that both the electron and the proton rotate about the centre of mass of the electron-proton system, rather than rotating about the centre of mass of the proton as assumed in the Bohr theory.
Thus, from (2-7) we find that the electron can have any one of a series of negative energies, which are referred to as energy states or levels. The lowest value of energy (the most negative) corresponds to setting n=1 in (2-7). Since this is the lowest energy, it is called the ground state. The actual value of the discrete negative energy indicates the amount of energy required to remove that electron completely away from its orbit around the nucleus, and thus that energy is sometimes referred to as a binding energy. It is also referred to as the ionization energy, since the result is to create a positively charged ion and a free electron. Once an electron is removed from the attractive force of the ion, it can have any value of velocity and thus can have any value of positive energy associated with its kinetic energy.
Example Calculate the binding energy of the electron for the first three energy levels of the hydrogen atom.