Early quantum theory and early models of the atom

Discovery of electron

In the same year, J. S. Townsend first measured the charge, $e$ of an electron. But the more accurate value of, $e$, was obtained by Robert Millikan in 1909 with his famous experiment called Millikan’s oil drop experiment. From the value of Thomson’s $e/m_e$ and the value of $e$, the mass of the electron, $m_e$, was determined.

Blackbody radiation

The total intensity, $I$ of radiation emitted by an object depends on the temperature. It is proportional to the fourth power of the absolute temperature as you learned in the topics on heat.

i.e., $I\propto T^4$.

At low temperature, around 300K (room temperature), the intensity of radiation is very small, therefore we cannot feel or see the radiation.

At higher temperatures but lower than 1000K, objects emit infrared radiation. The radiation can be felt near the object.

At temperatures above 1000K, objects glow. Examples: red-hot burner of an electric stove and heating element of a toaster.

At temperatures above 2000K, objects glow with yellow or whitish color. Example: filament of an incandescent light bulb.

Blackbody spectrum

Wien’s law

i.e., $\lambda_p \,T=2.90\times 10^{-3} \:m\,K$.

This equation is called Wien’s law.

Planck’s quantum hypothesis

The minimum energy of atomic or molecular oscillation (according to Planck's hypothesis) is $hf$, where $f$ is the frequency of the oscillation and $h$ is a constant called Planck's constant. The value of $h$ is

$h=6.626\times 10^{-34}Js$

$E=nhf$

The most notable point is, the energy of atomic or molecular oscillation is not continuous, but is discrete, i.e., the energy can be $hf$, $2hf$, $3hf$, .... and there is no other values in-between. Therefore, we say that the energy of atomic or molecular oscillation is quantized.

The Plank's quantum theory is the beginning of quantum physics.

Einstein photon theory of light

Energy of a photon is

$E=hf$

Momentum of a photon is

$p=\dfrac{h}{\lambda}$

Photoelectric effect

An apparatus used to demonstrate the photoelectric effect is the photocell. A photocell consists of a curved metal plate, called (emitter plate) and a small electrode, called collector plate in an evacuated glass enclosure. A battery and an ammeter (A) are connected in series to the photocell. The positive terminal of the battery is connected to the collector plate and the negative terminal to the emitter plate. So, the collector plate is at higher potential and the emitter plate is at lower potential. When light illuminates the emitter plate, it emits electrons. The emitted electrons move toward the collecting plate as electrons move from lower potential to higher potential. Due to the flow of electrons, a current flows in the circuit that is indicated by the ammeter. When no light incident on the plate, there is no current flows in the circuit.

The photoelectrons are emitted with a range of speeds, so the kinetic energies. The kinetic energy of the fastest moving electrons is known as the maximum kinetic energy. By reversing the polarity of the battery, and adjusting the voltage, we can find the maximum kinetic energy. When the voltage is reversed, i.e., connecting the negative terminal to the collecting plate and the positive terminal to the emitter plate, the collector plate is now at lower potential. So, the photoelectrons experience a repulsive force by the collector plate, which tries to stop the electrons. The repulsive force is proportional to the magnitude of the voltage across the plates. So, a certain voltage can stop the electrons of up to a certain speed. If we increase the voltage from 0 V, at one point there will be no current. At this point we have stopped all the electrons including the fastest moving.The voltage at which all the electrons are just stopped so that no current in the circuit is called the stopping voltage or the stopping potential.

Since the stopping potential is just the correct amount of voltage to stop the fastest moving electrons, we can relate it to the maximum kinetic energy of the electrons. If an electron moving with a kinetic energy, $KE_{max}$ is stopped, then the loss in kinetic energy by the electron is equal to the electric potential energy gained by the electron. At the stopping potential, $V_0$, the energy gained by the electron is $eV_0$, therefore, we have

$KE_{max}=eV_0$

Experiments on photoelectric effect reveal the following facts.

• There is no electrons emitted if the frequency of the light is below a certain frequency, called cutoff frequency.
• If the frequency of the light is above the cutoff frequency, there is always electrons emitted even if the intensity is low.
• If the frequency is above the cutoff frequency, increasing the intensity increases the number of photoelectrons, but it does not change the maximum kinetic energy of the electrons.

Electromagnetic wave theory of photoelectric effect

• Increasing the intensity of the light increases the maximum kinetic energy of the ejected electrons. This is because, increasing the intensity of the light increases the amplitude of the electric field, which makes the electrons on the metals oscillate violently so that they are ejected at higher velocities.
  
• Frequency of the light does not affect the maximum kinetic energy the electrons.

The above predictions by electromagnetic wave theory is in contrary to what is actually happening. So, we can conclude that the wave theory fails to explain the photoelectric effect.

Einstein theory of photoelectric effect

• It takes one electron to eject one photon, and during the process the photon is fully destroyed. Increasing the intensity of light, increases the number of ejected electrons as increasing the intensity increases the number of photons.
  
• Electrons in a metal are bound to the metal by a binding energy, which is called work function, $W_0$. If the energy of the photon is less than the work function, then no photoelectrons emitted. So, the minimum energy of photon for photoelectric effect is

  $hf_0=W_0$

  where $f_0$ is the cutoff frequency for the photoelectric effect.

• If the frequency of the light is greater than the cutoff frequency, then there are electrons emitted. And, above the cutoff frequency, if the frequency is increased, the maximum kinetic energy of the photoelectrons increases according to the following equation:

$KE_{max}=hf-W_0$

A graph of the maximum kinetic energy of photoelectrons vs the frequency of the light (photon) is given below.

Wave-particle duality

Matter waves

$\lambda=\dfrac{h}{p}$

Early models of the atom

• Plum pudding model and
• Rutherford planetary model.

Plum pudding model

The plum pudding model was developed in 1904 by J. J. Thomson. This model considered atom as a homogeneous sphere of positive charge with negatively charged electrons embedded in it. That is, electrons are like plums inside a pudding of a positive charge.

Rutherford planetary model

Atomic spectra of gases

The emission spectrum of the hydrogen gas is given below. There are four lines in the visible region with the wavelengths, 410 nm, 434 nm, 486 nm and 656 nm.

The emission spectrum is unique for a gas, so we can identify a gas by its spectrum.

Further, if we pass white light through a gas, and examine the transmitted light, we will see that certain wavelengths of the light are missing. If we pass the transmitted light through a diffraction grating, you will see the spectrum of the light with dark lines for the missing wavelengths. The light corresponding to the missing wavelengths are absorbed by the gas. The interesting thing is the missing wavelengths correspond to the wavelengths in the emission spectrum. i.e., a gas absorbs and emits light at the same wavelengths (or frequencies).

The hydrogen atom

You see that there are three groups of spectral lines, called Lyman series, Balmer series and Paschen series.

These series were named so because, Lyman, Balmer and Paschen found a formula to fit the spectral lines in the UV, visible and the IR spectral lines respectively.

J. J. Balmer in 1885, found a formula to fit the visible spectral lines of the hydrogen atom. Later, his formula was generalized to include other wavelengths in the series that we call Balmer series. The formula for the Balmer series is

$\dfrac{1}{\lambda}=R \bigg (\dfrac{1}{2^2}-\dfrac{1}{n^2}\bigg )$,   $n=3,4,5,...$

where $R$ is a constant called, Rydberg constant, $R=1.09737 \times 10^7 m^{-1}$ and $n= 3, 4, 5, .....$. If you put $n=3$, you will get, $\lambda = 656 nm$, and $n=4$, $\lambda = 486 \,nm$, $n=5$, $\lambda = 434 \,nm$ and $n=6$, $\lambda = 410 \,nm$, the wavelength of the four visible lines in the spectrum. The Balmer series covers other spectral lines of wavelength up to 365 nm. The 365 nm corresponds to $n=\infty$.

The Lyman sries covers the spectral lines of wavelength from 91 nm to 122 nm. The formula for the Lyman series is

$\dfrac{1}{\lambda}=R \bigg (\dfrac{1}{1^2}-\dfrac{1}{n^2}\bigg )$,   $n=2,3,4,...$

The Paschen series of the spectral lines covers wavelength from 820 nm to 1875 nm. The formula for the Paschen series is

$\dfrac{1}{\lambda}=R \bigg (\dfrac{1}{3^2}-\dfrac{1}{n^2}\bigg )$,   $n=4,5,6,...$

With the Rudherford model, we cannot explain the spectrum of an atom. So, we need another model of the atom. In 1913, Niels Bohr developed a model of the atom. His model was successful in explaining the emission spectrum of the hydrogen atom.

The Bohr model of atom

1. electrons move in circular orbits around the nucleus but only certain orbits are allowed. An electron in each orbit has a definite energy, and when an electron moves in its orbit, it does not radiate energy. Note that, if an object is in circular motion, then it has an acceleration. So, the electron in an orbit has an acceleration. Since an accelerating charge emits radiation, the electron should also emits radiation. If it does, then it loses its energy and finally would collapse into the nucleus, but this doesn't happen. So, Bohr theorized that when an electron revolving around the nucleus in an orbit, it does not radiate energy. The orbits of electrons are called stationary states.

When an electron jumps from a stationary state of higher energy (upper state) to another stationary state of lower energy (lower state), it loses energy. This lost energy is emitted as a photon. The energy of the emitted photon is exactly equal to the energy difference between the two states. That is,

$hf=E_u-E_\ell$

2. Angular momentum of an electron in an atom is quantized. That is an electron can have only certain discrete values of angular momentum. According to Bohr, the angular momentum of an electron in an orbit is

$L=n\dfrac{h}{2\pi}$

where $n$ is the orbit’s number, and is called principal quantum number.

Hydrogen atom's orbits

The angular momentum of the electron in the orbit is

$L=mr_nv$

where $m$ is the mass of the electron, and $v$ is the velocity of the electron in the orbit.

From the previous section, we have,

$L=n\dfrac{h}{2\pi}$

From the above two equations, we can find the speed of the electron in an orbit:

$v=\dfrac{nh}{2\pi r_n m}$

Since the nucleus of the atom is positively charged, there is a force of attraction between the nucleus and the electron. The charge of nucleus of the hydrogen atom is $+e$ as there is only one proton in the nucleus. So, the magnitude of the force according to the Coulomb's law is

$F= k \dfrac{e\cdot e}{r_n^2}$

This force keeps the electron in its circular orbit. So, this is the centripetal force on the electron. Therefore,

$\dfrac{mv^2}{r_n}=k \dfrac{e^2}{r_n^2}$

Substituting $v$ from above, and solving for $r_n$, we get

$r_n=\dfrac{n^2h^2}{4 \pi^2 m k e^2}$

Now, substituting the values of $h, m, k$ and $e$, we will get,

$r_n=(0.529 \times 10^{-10})\, n^2$

This is the radius of the n^th orbit that is proportional to n².

For the first orbit, $n=1$, putting $n=1$ in the above equation, you will get the radius of the first orbit:

$r_1=0.529 \times 10^{-10}\, m$

This is the radius of the smallest orbit of the electron.

The radius of the second ($n=2$), orbit is

$r_2=2.12 \times 10^{-10}\, m$

And, for the third orbit (n=3), the radius is

$r_3=4.76 \times 10^{-10}\, m$

Energy of the electron orbits

Now, let us find the energy of the orbits. The energy of an orbit is the total energy of an electron in that orbit. An electron in an orbit has potential energy and kinetic energy.

The potential energy of the electron is

$PE=qV=-eV$

where $V$ is the electric potential at the position of the electron. This electric potential is due to the nucleus of the atom. When the electron is in the n^th orbit, it is at a distance of $r_n$ from the nucleus. So, the electric potential due to the nucleus is

$V=\dfrac{kq}{r_n}=\dfrac{ke}{r_n}$.

Substituting this in the above equation, you get,

$PE=-\dfrac{ke^2}{r_n}$

The kinetic energy of the electron is

$KE=\dfrac{1}{2}mv^2$

So, the total energy of the electron in the n^th orbit is

$E_n=\dfrac{1}{2}mv^2-\dfrac{ke^2}{r_n}$

Now, substituting, $v$ and $r_n$ from the previous section, and simplifying, we get

$E_n=-\dfrac{2 \pi^2e^4mk^2}{h^2 n^2}$,   

Subsitituting the values of the constants, $e$, $m$, $k$ and $h$, we get

$E_n=-\dfrac{13.6 \,eV}{n^2}$,     $n=1, 2 , 3, ...$

The energy of first orbit is

$E_1=-13.6 \, eV$

This is the energy of the lowest state. This state is called the ground state.

The energy of the second (n=2) and the third (n=3) states are

$E_2= -3.40 \,eV$

$E_3=-1.51 \,eV$

So the energy is not continuous but is discrete. Because, we have energies, -13.6 eV, -3.40 eV, -1.51 eV,..., and there is no energy in between these values. That is the energy is quantized. Note that the energy is negative. The orbits and the energy of each orbit are shown in the figure below.

Spectral lines and the Bohr model

As shown in the figure, each spectral line in the Lyman series is due to the jumping of electrons from one of the higher state to the ground state (n=1). In Palmer series, each spectral line is the result of electrons jumping from a higher state to the n=2 state. And, each line in the Paschen series is due to the electrons jumping from one of the higher state to the n= 3 state.

Now, let us consider an electron make a transition from a state n to a lower state n'. During this process a photon is released. The energy of the emitted photon is

$hf=E_n-E_{n'}$

We have, $E_n=-\dfrac{2 \pi^2e^4mk^2}{h^2 n^2}$

Substituting, $f=c/\lambda$ and $E_n$ and $E_{n'}$ from the above equation, we get

$\dfrac{hc}{\lambda}=-\dfrac{2 \pi^2e^4mk^2}{h^2 n^2}+\dfrac{2 \pi^2e^4mk^2}{h^2 n'^2}$

From this, we can write,

$\dfrac{1}{\lambda}=\dfrac{2 \pi^2e^4mk^2}{h^3c}\bigg(\dfrac{1}{n'^2}-\dfrac{1}{n^2}\bigg)$

Now, substituting the constants, we get

$\dfrac{1}{\lambda}=R\bigg(\dfrac{1}{n'^2}-\dfrac{1}{n^2}\bigg)$

This equation is of the same form as the formula we used for the Lyman, Balmer or the Paschen series with n'=1 corresponds to the Lyman series, n'=2, the Balmer series and n'=3, the Paschen series. So, with the Bohr theory we have derived those formulae. Thus the Bohr model of atom is successful in explaining the spectrum of hydrogen atom.