# Bohr Model of Atom

In the Bohr model of the hydrogen atom, an electron orbits a proton (the nucleus) in a circular orbit of radius $0.53*10^{-10} m$.

a. What is the electric potential at the electron’s orbit due to the proton?

b.What is the kinetic energy of the electron? Express your answer in joules ?

c.What is the kinetic energy of the electron? Express your answer in eV?

d.What is the total energy of the electron in its orbit? Express your answer in joules.

e. What is the ionization energy-that is, the energy required to remove the electron from the atom and take it to $r = \infty$ , at rest? Express your answer in joules

In the Bohr model, the angular momentum $L=r\times p$ ($\times$ is the vector product) vector is quantized.

$L= n\hbar$

where $\hbar$ is the Planck reduced constant.

From this, and from the fact that the velocity of the electron comes from equating the centripetal force with the coulomb force

$m*v^2/r = k*e^2/r^2$ , ($k$ is here the electrostatic constant, $m$ is the mass of electron)

the value of orbit diameters are

$r(n) = (n^2*\hbar^2)/(k*e^2*m)$

and the smallest possible orbit is therefore $r(1) = 0.53*10^{-10} m$ for the principal quantum number $n=1$

a) the electric potential is simply

$V(r) = E(r)*r = F(r)*r/e = (k*e^2/r^2) *r/e =k*e/r =$

$=9*10^9*1.6*10^-{19}/0.53*10^{-10} =27.17 V$

b) the speed comes from the quantization rule or from making equal the centripetal force with the coulomb force

$m*v^2/r = k*e^2/r^2$

$E c = m*v^2/2 = k*e^2/(2*r) =9*10^9*(1.6*10^-{19})^2/2/0.53*10^{-10}= 2.17*10^{-18} J$

c) In electron-volts the kinetic energy is

$E c (eV) = E c (J)/e =2.17*10^{-18}/1.6*10^{-19} = 13.58 eV$

d) In the Bohr atom model, the energy on the n level is given by

$E(n) = -k*e^2/(2*r(n))$

and when comparing with the energy at point b) one can say that the energy of the level is the kinetic energy taken with the sigh changed (BY DEFINITION)

hence $E(1) = -13.6 eV =-2.17*10^{-18} J$

The total energy at any radius comes from

$E(n)= Ek + Epot = m*v^2/2 + k*e^2/r = -k*e^2/(2*r)$

This is why the TOTAL energy is equal to the kinetic energy taken with the sign changed.

To take the electron from this energy to infinity at rest (or in other words to zero energy) it would take the energy $-E(1)$.

Hence the first ionization energy is $E = 2.17*10^{-18} J$