Photoelectric Effect

Notes for Photoelectric Effects chapter of class 12 physics. Dronstudy provides free comprehensive chapterwise class 12 physics notes with proper images & diagram.

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Photoelectric effect

The emission of electrons from a metallic surface when irradiated by electromagnetic radiation is called the phenomenon of photoelectric effect. The emitted electrons are called asÂ photoelectrons
For the investigation of the photoelectric effect a schematic diagram of the apparatus as used by Lenard (1902) is shown in the figure. Monochromatic light from the lamp L illuminates a plate P in an evacuated glass enclosure. A battery maintains a potential difference between P and a metal cylinder C, which collects the photoelectrons. The potential C can be varied to be either positive or negative relative to P.

When the collector is positive with respect to the plate, the electrons are attracted to it and the ammeter (A) registers a current. Lenard studied the dependence of photoelectric current on the following factors.
(i)Â Â  Intensity of incident radiation
(ii)Â  Potential difference between the plate and the collector
(iii) Frequency of the incident radiation
The result of observations are as follows:

Effect of the intensity of Incident Radiation

When the collector is positive relative to the plate and the potential difference is kept fixed, then for a given frequency of radiation, the photoelectric current is proportional to the intensity of the light, as shown in the figure. Â It shows that the number of emitted photoelectrons is proportional to the light intensity. Furthermore, there is no threshold intensity.

Effect of Potential Difference

When the frequency and intensity of radiation are kept constant and the positive potential of collector relative to plate is gradually increased, then the photoelectric current i increases with the potential difference V. At some value of the potential difference, when all the emitted electrons are collected, thus increasing potential difference has no effect on the current. The current has reached its maximum value, called the saturation current.
When the polarity of the battery is reversed, the electrons are repelled and only the most energetic ones reachÂ  the collector, so the current falls. When the retarding potential difference reaches a critical value, the current drops to zero. At this stopping potential Vo, only those electrons with the maximum kinetic energy are able to reach the collector.
$e{V_0} = {1 \over 2}mv_{\max }^2$
For a given frequency of light, the saturation current depends on the intensity of light. Larger the intensity; higher the saturation current. However, the stopping potential does not change with the intensity. It is clearly shown in figure.

Effect of frequency

For a given intensity of radiation, the stopping potential depends on the frequency. Higher the frequency, higher the value of stopping potential.
The maximum kinetic energy of the electrons depends on the light source and the plate material, but not on the intensity of the source. Certain combinations of light sources and plate materials exhibit no photoelectric effect.

Einsteinâ€™s Theory of Photoelectric Effect

According to Einstein, the experimental results of photoelectric effect can be explained by applying the quantum theory of light. He assumed that light of frequency n contain packets or quanta of energy E = hn. On this basis, light consists of particles, and these arecalled photons. The number of photons per unit area of cross-section of the beam of light per second is proportional to its intensity. But the energy of photon is proportional to its frequency and is independent of the light intensity.
In the process of photo emission a single photon gives up all its energy to a single electron. As a result, the electron is ejected instantaneously. Since the intensity of light is determined by the number of photons incident, therefore, increasing the intensity will increase the number of ejected electrons.
The maximum possible kinetic energy$\left( {{1 \over 2}mv_{max}^2} \right)$of the photoelectrons is determined by the energy of each photon (hn) according to the Einstein equation,

${1 \over 2}mv_{\max }^2 = hv - W$

where the work function, (W), is the minimum energy needed to extract an electron from the surface of the material.
In terms of threshold frequency, it is given by

$W{\rm{ }} = {\rm{ }}h{n_o}$

Using the above equation, we may write the Einstein equation as
${\textstyle{1 \over 2}}m{v^2}_{\max } = hn-{\rm{ }}h{n_o} = {\rm{ }}h(n-{n_o})$
Also, in terms of stopping potential,

$e{V_o} = h(n-{n_o})$

Application 1

Ultraviolet light of wavelength 2000Ã… causes photoemission from a surface. The stopping potential is 2V.
(a)Find the work function in eV
(b)Find the maximum speed of the photoelectrons.

solution

a)Using Einstein relation
$W = {\textstyle{{hc} \over \lambda }} - e{V_0}$
or $W = {{12400} \over {2000}} - 2 = 4.2eV$
b)Since ${1 \over 2}mv_{\max }^2 = e{V_o}$
${v_{\max }} = \sqrt {{{2e{V_o}} \over m}} = \sqrt {{{2\left( {1.6 \times {{10}^{ - 19}}} \right)\left( 2 \right)} \over {9.1 \times {{10}^{ - 31}}}}}$
orÂ Â  ${v_{\max }} = 8.4 \times {10^5}m/s$

Determination of photoelectric current

Let P be the power of a point source of electromagnetic radiations, then intensity I at
a distance r from the source is given by
$I = {p \over {4\pi {r^2}}}(W/{m^2})$
If A is the area of a metal surface on which radiations are incident, then the power received by the plate is
$p' = IA = \left( {{P \over {4\pi {r^2}}}} \right)A(W)$
If n is the frequency of radiation, then the energy of photon is given by
$E = hv$
The number of photons incident on the plate per second (called photon flux) is given by
$\Phi = {{P'} \over E} = \left[ {{{{P \over {4\pi {r^2}}} \times A} \over {h\nu }}} \right]$

If n > n o (threshold frequency) and photon efficiency of the metal plate is $\eta \%$ then the number of photoelectrons emitted per second is given by
$n = {{\Phi \eta } \over {100}} = \left[ {{{{P \over {4\pi {r^2}}} \times A} \over {h\nu }}} \right]{\eta \over {100}}$
Finally, the photocurrent i is given by
$i = ne$
where e is the charge of an electron
$(e = 1.6 \times {10^{ - 19}}C)$

Application 2

The intensity of sunlight on the surface of earth is 1400 W/m2Â Assuming the mean wavelength of sunlight to be 6000Ã…, calculate
(a)the photon flux arriving at 1 m2 area on earth perpendicular to light radiations, and
(b)the number of photons emitted from the sun per second assuming the average radius of Earthâ€™s orbit is $1.49 \times {10^{11}}m$.

SolutionÂ

(a)Energy of a photon$E = {{hc} \over \lambda } = {{12400} \over {6000}} = 2.06\,eV = 3.3 \times {0^{ - 19}}J\,$
Photon flux Â =${{IA} \over E} = {{\left( {1400} \right)\left( 1 \right)} \over {3.3 \times {{10}^{ - 19}}}} = 4.22 \times {10^{21}}$
(b)$n = {P \over E} = {{I\left( {4\pi {R^2}} \right)} \over E}$

or $n = {{\left( {1400} \right)\left( {4\pi } \right){{\left( {1.49 \times {{10}^{11}}} \right)}^2}} \over {3.3 \times {{10}^{ - 19}}}} = 1.18 \times {10^{45}}$

Matter Waves

According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.

or

A wave is associated with moving material particle which control the particle in every respect.

The wave associated with moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with group velocity.

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de-Broglie wavelength

According to de-Broglie theory, the wavelength of de-Broglie wave is given by

$\lambda = {h \over p} = {h \over {mv}} = {h \over {\sqrt {2mE} }}$Â $\Rightarrow \lambda \propto {1 \over p} \propto {1 \over v} \propto {1 \over {\sqrt E }}$ Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â Â

Where h = Plank's constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle.

The smallest wavelength whose measurement is possible is that of -rays.

The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, a-particle etc. is of the order 10-10 of m.Â

de-Broglie wavelength associated with the charged particles.

The energy of a charged particle accelerated through potential difference V isÂ $E = {1 \over 2}m{v^2} = qV$

Hence de-Broglie wavelengthÂ $\lambda = {h \over p} = {h \over {\sqrt {2mE} }} = {h \over {\sqrt {2mqV} }}$

${\lambda _{electron}} = {{12.27} \over {\sqrt V }}$Ã…,Â Â  ${\lambda _{proton}} = {{0.286} \over {\sqrt V }}$Ã…,Â Â Â Â  ${\lambda _{deutron}} = {{0.202 \times {{10}^{ - 10}}} \over {\sqrt V }}$Ã…,Â Â  ${\lambda _{\alpha - particle}} = {{0.101} \over {\sqrt V }}$Ã…Â

de-Broglie wavelength associated with uncharged particles.

For Neutron de-Broglie wavelength is given as ${\lambda _{Neutron}} = {{0.286 \times {{10}^{ - 10}}} \over {\sqrt {E\,({\rm{in}}\,eV)} }}m = {{0.286} \over {\sqrt {E\,({\rm{in}}\,eV} )}}$Ã…

Energy of thermal neutrons at ordinary temperature

$E = kT \Rightarrow \lambda = {h \over {\sqrt {2mkT} }}$

Â ;Â  where k = Boltzman's constant = Joules/kelvin , T = Absolute temp.

So ${\lambda _{{\rm{Thermal}}\;{\rm{Neutron}}}} = {{6.62 \times {{10}^{ - 34}}} \over {\sqrt {2 \times 1.07 \times {{10}^{ - 17}} \times 1.38 \times {{10}^{ - 23}}T} }} = {{30.83} \over {\sqrt T }}$Ã…

Characteristics of matter waves

(i) Matter wave represents the probability of finding a particle in space.

(ii) Matter waves are not electromagnetic in nature.

(iii) de-Brogile or matter wave is independent of the charge on the material particle. It means, matter wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).

(iv) Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles is nature. Â

(v) Electron microscope works on the basis of de-Broglie waves.

(vi) The electric charge has no effect on the matter waves or their wavelength.

(vii) The phase velocity of the matter waves can be greater than the speed of the light.

(viii) Matter waves can propagate in vacuum, hence they are not mechanical waves.

(ix) The number of de-Broglie waves associated with nth orbital electron is n.

(x) Only those circular orbits around the nucleus are stable whose circumference is integral multiple of de-Broglie wavelength associated with the orbital electron.Â Â Â

DAVISION AND GERMER EXPERIMENT

It is used to study the scattering of electron from a solid or to verify the wave nature of electron. A beam of electrons emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. Ni crystal behaves like a three dimensional diffraction grating and it diffracts the electron beam obtained from electron gun.

Â The diffracted beam of electrons is received by the detector which can be positioned at any angle by rotating it about the point of incidence. The energy of the incident beam of electrons can also be varied by changing the applied voltage to the electron gun.

According to classical physics, the intensity of scattered beam of electrons at all scattering angle will be same but Davisson and Germer, found that the intensity of scattered beam of electrons was not the same but different at different angles of scattering.

Â Intensity is maximum at 54 V potential difference and 50o diffraction angle.

If the de-Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg's formula $2d\sin \theta = n\lambda$, we can determine the wavelength of these waves.

Where d = distance between diffracting planes, $\theta = {{(180 - \varphi )} \over 2}$= glancing angle for incident beam = Bragg's angle.

The distance between diffraction planes in Ni-crystal for this experiment is d = 0.91Ã… and the Bragg's angle = 65o. This gives for n = 1,Â $\lambda = 2 \times 0.91 \times {10^{ - 10}}\sin {65^o} = 1.65$Â Ã…

Now the de-Broglie wavelength can also be determined by using the formula:

$\lambda = {{12.27} \over {\sqrt V }} = {{12.27} \over {\sqrt {54} }} = 1.67$Â Â Ã…

Thus the de-Broglie hypothesis is verified.

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