Atomic spectra, The Bohr's atom

Atomic spectra:

Atomic stability is not the only thing that a successful theory of atom must account for. The existence of spectral lines is another important aspect of the atom.

When an atomic gas or vapour at low pressure is suitably excited, usually by passing an electric current through it, the emitted radiation has a spectrum which contains certain specific wavelength only.

The number, intensity and exact wavelength, of the lines in the spectrum of an element depend upon temperature, pressure, the pressence of electric and magnetic fields and the motion of the source. It is possible to tell by examining its spectrum not only what element are present in a light source but much about their physical state.

Bohr's atomic model:

Assumption

1. Electron in an atom revolves in a circular order. During the motion of electron it doesn't radiate energy. The radiation of orbit is constant, such orbits are known as stationary orbit.

2. A angular momentum of electron revolving around nucleus is intrigal multiple of $\frac{n}{2\pi}$. $$angular\; momentum= mvr=\frac{nh}{2\pi}$$ $$or,\;\;mvr=n\hbar$$ $$where, \;\;\;\hbar=\frac{h}{2\pi}$$ $$h= Plank's constant$$

Equation (1) indicates that, angular momentum in an atom is discrete ( or quantized)

3. When an electron jumps from higher energy level to lower energy level then it emits energy in the form of electromagnetic radiation ( photon ) wherever when electron jumps from lower energy level to higher energy level it absorbs energy in the form of radiation.

$$i.e,\;\; h\nu=E_i-E_f \;\;\;\; [E_i<E_f]$$ $$\Rightarrow \nu=\frac{E_i-E_f}{h}\dotsm(2)$$ Equation (2) gives the frequency of radiation emmited when electron jumps from level having energy $E_i$ to another level having energy $E_f$.

Bohr's theory of hydrogen atom:

Consider an electron of mass M charge of magnitude e, revolving around nucleus in an circular orbit of radius r as shown in figure.

Figure here

The centripetal force on electron is due to electrostatic force between nucleus and electron.

ie. $$\begin{align*} \frac{mv^2}{r}= \frac{1}{4\pi\epsilon_\circ}\frac{(Ze)e}{r^2}\dotsm(3)\\ \\Where, Z= Atomic number\\ v= Speed of electron\\ Ze = Charge of nucleus\end{align*}$$

$$\begin{align*} (mvr)V= \frac{1}{4\pi\epsilon_\circ} Ze^2 \end{align*}$$ using equation (1) $$\begin{align*} n\frac{h}{2\pi}V=\frac{(Ze)^2}{4\pi\epsilon_\circ}\end{align*}$$ $$\begin{align*}V_n= \frac{Ze^2}{2n\epsilon_{\circ}h}\dotsm(4)\end{align*}$$

Equation (4) gives the speed of a electron in $n^{th}$ orbit

$$\begin{align*}i.e V_n\propto Z\end{align*}$$

$$\begin{align*} V_n\propto \frac{1}{n}\end{align*}$$

Again from equation (3) $$\begin{align*} r= \frac{ze^2}{4\pi\epsilon_\circ}\frac{1}{mv^2}\end{align*}$$ Substituting equation(4) in above equation, we get: $$\begin{align*} r=\frac{Ze^2}{4\pi\epsilon_\circ}\frac{4n^2\epsilon_\circ^2h^2}{mZe^4}\end{align*}$$ $$\begin{align*}r_n=\frac{n^2h^2\epsilon_ \circ}{{\pi}mZe^2}\dotsm(5)\end{align*}$$

Equation (5) give the radious of $n^{th}$ orbit.

$$\begin{align*} or, r_n= n^2\biggl[\frac{h^2\epsilon_\circ}{{\pi}m2e^2}\biggr]\end{align*}$$ $$\begin{align*} =n^2\biggl[\frac{(6.625 X 10^{-34})^2 X (8.853 X 10^{-12})}{3.14 X 9.1 X 10^{-31} X 2( 1.602 X 10^{-19})^2}\biggr]\end{align*}$$

$$\begin{align*} r_n= 0.529 X 10^{-10} \frac{n^2}{2}\dotsm(6)\end{align*}$$

$$\begin{align*} r_n\propto n^2\;\; and\; r_n\propto \frac12\end{align*}$$

The radious of first electronic orbit is given by $$r_1= 0.529 X 10^{-10}M \; ( H- atom)$$ $$r_1= 0.529A^{\circ}$$

Calculation of total energy of electron in H-atom

The total energy is the sum of kinetic energy and potential energy of electron $$i.e\;\; E= K.E + P.E$$ $$E= \frac12MV^2+\frac{1}{4\pi\epsilon_\circ}\frac{(+Ze)(-e)}{r}$$ using equation (3) we get: $$E=\frac12\biggl(\frac{Ze^2}{4\pi\epsilon_{\circ}r}\biggr)-\biggl(\frac{Ze^2}{4\pi\epsilon_{\circ}r}\biggr)$$

$$=\frac{-Ze^2}{8\pi\epsilon_ {\circ}r}$$ Using eqation (5), we get

$$E_n=\frac{-Ze^2}{8\pi\epsilon_\circ}\frac{{\pi}mZe^2}{n^2h^2\epsilon_\circ}$$

$$E_n= \frac{-mZ^2e^4}{8{\epsilon_\circ}^2n^2h^2}\dotsm(7)$$

Equation (7) gives the energy of electron in $n^{th}$ orbit by putting value of all constant term we ge,

$$E_n= -13.6ev \frac{z^2}{n^2}\;\;\;\;\;\; (n= 1,2,3, . , . ,\infty)$$ $$E_infty= \frac{-13.6}{\infty^2}=0 \;ev$$

we see that electron at infinity has zero energy. This means that an electron at infinity orbit in an atom is almost free from it.

For H-atom, Z=1 $$E_n=\frac{-13.6ev}{n^2}$$ At ground state (Minimum energy state) n=1 $$E_1=-13.6ev$$ for first exited state, n=2 $$E_2=\frac{-13.6ev}{2^2}=\frac{-13.6ev}{4}=-3.4ev$$

similarly, for second excited state $E_3=1.51ev$ $E_4= -0.85 ev \;and \;so\; on\; for \;n=\infty$, $E_\infty$= 0

Thus this indicate the energy level of H-atom is discrete at low energy.

Reference:

Reviews of Modern Physics. Lancaster, P.A.: Published for the American Physical Society by the American Institute of Physics, 1952. Print.

Wehr, M. Russell, and James A. Richards. Physics of the Atom. Reading, MA: Addison-Wesley Pub., 1984. Print.

Young, Hugh D., and Roger A. Freedman. University Physics. Boston, MA: Pearson Custom, 2008. Print.

Adhikari, P.B. A Textbook of Physics. 2070 ed. Vol. II. Kathmandu: Sukunda Publication, 2070. Print.