1.8 Experimental Study of Photo Electric Effect :
Experiments with the photoelectric effect are performed in a discharge tube apparatus as illustrated in figure shown. The cathode of discharge tube is made up of a metal which shows photoelectric effect on which experiment is being carried out.
A high potential is applied to a discharge tube through a variable voltage source and a voltmeter and an ammeter are connected a measure the potential difference across the electrodes and to measure photoelectric current. Light with frequency more than threshold frequency of cathode metal is incident on it, due to which photoelectrons are emitted from the cathode. These electrons will reach the anode and constitute the photoelectric current which the ammeter will show.
Now we start the experiment by closing the switch S. Initially the variable battery source is set at zero potential. Even at zero potential variable source, ammeter will show some current because due to the initial kinetic energy some electrons will reach the anode and cause some small current will flow. But as we know majority of ejected electrons have low values of kinetic energies which are collected outside the cathode and create a could of negative charge, we call space charge, as shown in figure shown.
If the potential difference applied across the discharge tube is gradually increased from the variable source, positive potential of anode starts pulling electrons from the space charge. As potential difference increases, space charge decrease and simultaneously the photoelectric current in circuit also increases. This we can also see in the variation graph of current with potential difference as shown in figure shown.
A shown in graph, we can see as potential difference increases, current in circuit increases. But at a higher voltage V_{P1} space charge vanishes and at this voltage anode is able to pull the slowest electron (zero kinetic energy) ejected by the cathode. Now as all the ejected electrons from cathode start reading anode. If further potential difference is increased, it will not make any difference in the number of electrons reaching the anode hence, further increases in potential difference will not increases the current. This we can see in figure shown that beyond V_{P1} current in circuit becomes constant. This current i_{s1} is called saturation current. This potential difference V_{P1} at which current becomes saturated is called "pinch off voltage".
Now if the frequency of incident light is kept constant and its intensity is further increased, then the number of incident photons will increase which increases the number of ejected photo electrons so current in circuit increases and now in this case at higher intensity of incident light, current will not get saturated at potential difference V_{P1} as now due to more electron emission, space charge will be more and it will not vanish at V_{P1}. To pull all the electrons emitted from cathode more potential difference is required. This we can se from figure shown, that at higher intensity I_{2} (I_{2} > I_{1}) current becomes saturated at higher value of potential difference V_{P2}.
Intensity I_{2} > I_{1}
Beyond V_{P2}, we can see that all the electrons ejected from cathode are reaching the anode are current become saturated at i_{s2} because of more electrons. Another point we can see from figure shown that when V = 0 then also current is more at high intensity incident radiation as the number of electrons of high kinetic energy are also more in the beginning which will reach anode by penetrating the space charge.
1.9. Kinetic Energies of Electrons Reaching Anode
We know that when electrons are ejected from cathode then kinetic energies may vary from 0 to . If V is the potential difference applied across the discharge tube then it will accelerates the electron while reaching the anode. the electron which is ejected from cathode with zero kinetic energy will be the slowest one reaching the anode if its speed is v_{1} at anode then we have
0 + ve =
Similarly the electron ejected from cathode with maximum kinetic energy will be the fastest one when it will reach anode. If its speed is v_{2} at anode then we have
Thus we can say that all the electrons reaching anode will have their speeds distributed from v_{1} to v_{2}.
1.10 Reversed Potential Across Discharge Tube :
Now the experiment is repeated with charging the polarity of source across the discharge tube. Now positive terminal of source is connected to the cathode of discharge tube. When a light beam incident on the cathode with (hv > φ ), photoelectrons are ejected and move towards anode with negative polarity.
Now the electrons which are ejected with very low kinetic energy are attracted back to the cathode because of its positive polarity. Those electrons which have high kinetic energies will rush toward, anode and may constitute the current in circuit.
In this case the fastest electron ejected from cathode will be retarded during its journey to anode. As the maximum kinetic energy just after emission at cathode is , if potential difference across the discharge tube is V then the seed v_{f} with which electrons will reach anode can be given as
....(1)
Thus all the electrons which are reaching anode will have speed less then or equal to v_{f}. Remaining electrons which have relatively low kinetic energy will either be attracted to cathode just after ejection or will return during their journey from cathode to anode. Only those electrons will case current of flow in circuit which have high kinetic energies more then eV which can overcome the electric work against electric forces on electron due to opposite polarity of source.
1.11 Cut off Potential or Stopping Potential :
We have seen with reverse polarity electrons are retarded in the discharge tube. If the potential difference is increased with reverse polarity, the number of electrons reaching anode will decrease hence photo electric current in circuit also decreases, this we can see from figure shown which shows variation of current with increase in voltage across discharge tube in opposite direction. Here we can see that at a particular reverse voltage V_{0}, current in circuit becomes zero. This is the voltage at which the faster electron from cathode will be retarded and stopped just before reaching the anode.
Intensity I_{2} > I_{1} Frequency v (same for both radiation)
This voltage V_{0}, we can calculate from equation (1) by substituting v_{f} = 0 hence
 eV_{0} = 0
or eV_{0} =
or ...(2)
or ...(3)
We can see one more thing in figure shown that the graphs plotted for two different intensities I_{1} and I_{2}, V_{0} is same. Current in both the cases in cut off at same reverse potential V_{0}. The reason for this is equation(2) and (3). It is clear that the value of V_{0} depends only on the maximum kinetic energy of the ejected electrons which depends only on frequency of light and not on intensity of light. Thus in above two graphs as frequency of incident light is same, the value of V_{0} is also same. This reverse potential difference V_{0} at which the fastest photoelectron is stopped and current in he circuit becomes zero is called cut off potential or stopping potential.
1.12 Effect of Change in Frequency of Light on Stopping Potential :
If we repeat the experiment by increasing the frequency of incident light with number of incident photons constant, the variation graph of current with voltage will be plotted as shown in figure shown.
Frequency (v_{2} > v_{1})
This graph is plotted for two incident light beams of different frequency v_{1} and v_{2 }and having same photon flux. As the number of ejected photoelectrons are same in the two cases of incident light here we can see that the pinch off voltage V_{01} as well as saturation current i_{s1} are same. But as in the two cases the kinetic energy of fastest electron are different as frequencies are different, the stopping potential for the two cases will be different. In graph II as frequency of incident light is more, the maximum kinetic energy of photoelectrons will also be high and to stop it high value of stopping potential is needed. These here V_{01} and V_{02} can be given as
...(4)
and ...(5)
In general for a given metal with work function φ, if V_{o} is the stopping potential for an incident light of frequency v then we have
ev_{o} = hv  φ
or ev_{o} = hv  hv_{th}...(6)
or ...(7)
Equation (7) shows that stopping potential V_{0} is linearly proportional to the frequency v of incident light. The variation of stopping potential with frequency v can be shown in figure shown. Here equation (6) can be written as
= ev_{o}  h(v  v_{th})...(8)
This equation (8) is called Einstein's Photo Electric Effect equation which gives a direction relationship between the maximum kinetic energy stopping potential frequency of incident light and the threshold frequency.
Ex. 6 Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3 V. The photo electric effect begins in this metal at frequency of . Find the work function for this metal.
Sol. The threshold frequency for the given metal surface is
v_{th} = 6 X10^{14} Hz
Thus the work function for metal surface is
φ = hv_{th} = 6.63 X 10 ^{34 } X 6 X10^{14 }= 3.978 X 10^{19} J
As stopping potential for the ejected electrons is 3V, the maximum kinetic energy of ejected electrons will be
KE_{max} = 3eV = 3 X 1.6 X10 ^{19} J = 4.8 X 10 ^{19} J
According to photo electric effect equation, we have
hv = hv_{th }= KE_{max}
or frequency of incident light is
= 1.32 X10^{15} Hz
Ex.7 Electrons with maximum kinetic energy 3eV are ejected from a metal surface by ultraviolet radiation of wavelength 1500 Å. Determine the work function of the metal, the threshold wavelength of metal and the stopping potential difference required to stop the emission of electrons.
Sol. Energy of incident photon in eV is
According to photo electric effect equation, we have
or = 8.29  3 eV or = 5.29 eV
Threshold wavelength for the metal surface corresponding to work function 5.29 eV is given as
Stopping potential for the ejected electrons can be given as
Ex.8 Calculate the velocity of a photoelectron, if the work function of the target material is 1.24 eV and the wavelength of incident light is 4360 Å. What retarding potential is necessary to stop the emission of the electrons ?
Sol. Energy of incident photons in eV on metal surface is
= 2.85 eV
According to photo electric effect equation we have
or
= 2.85  1.24 eV = 1.61 eV
The stopping potential for these ejected electrons can be given as
Ex.9 Determine the Planck's constant h if photoelectrons emitted from a surface of a certain metal by light of frequency 2.2 × 10^{15} Hz are fully retarded by a reverse potential of 6.6 V and those ejected by light of frequency 4.6 × 10^{15} Hz by a reverse potential of 16.5 eV.
Sol. From photo electric effect equation, we have
Here hv_{1 } = φ + 2 eV_{01} ...(1)
and hv_{2} = φ + 2 eV_{02} ...(2)
Subtracting equation (1) from equation (2), we get
or
or or = 6.6 × 10^{34} Js
Ex.10 When a surface is irradiated with light of wavelength 4950 Å, a photo current appears which vanishes if a retarding potential greater than 0.6 volt is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. Find the work function of the emitting surface and the wavelength of second source. If the photo electrons (after emission from the surface) are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials.
Sol. In first case the energy of incident photon in eV is
= 2.51 eV
The maximum kinetic energy of ejected electrons is
KE_{max1} = eV_{01} = 0.6 eV
Thus work function of metal surface is given as
φ = E_{1}  KE_{max1} = 2.51  0.6 eV = 1.91 eV
In second case the maximum kinetic energy of ejected electrons will become
KE_{max2} = eV_{02 }= 1.1 eV
Thus the incident energy of photons can be given as
E_{2} = φ + KE_{max2}
E_{2} = 1.91 + 1.1 eV = 3.01 eV
Thus the wavelength of incident photons in second case will be
When magnetic field is present there will be no effect on the stopping potential as magnetic force can not change the kinetic energy of ejected electrons.
Ex.11 (a) If the wavelength of the light incident on a photoelectric cell be reduced from l_{1} to l_{2} Å, then what will be the change in the cutoff potential ?
(b) Light is incident on the cathode of a photocell and the stopping voltages are measured from light of two difference wavelengths. From the data given below, determine the work function of the metal of the cathode in eV and the value of the universal constant hc/e.
Wavelength (Å) Stopping voltage (volt)
4000 1.3
4500 0.9
Sol. (a) Let the work function of the surface be f. If v be the frequency of the light falling on the surface, then according to Einstein's photoelectric equation, the maximum kinetic energy KE_{max} of emitted electron is given by
We know that, KE_{max} = eV_{0}
Where V_{0}= cutoff potential.
Now, ΔV_{0 }= V_{02}  V_{01}
...(1)
(b) From equation (1), we have
= 1.44 × 10^{6} V/m
Now,
or
or φ = 2.3 eV
Ex.12 A low intensity ultraviolet light of wavelength 2271 Å irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3 V, find the work function of the metal. Will the photocell work if it is irradiated by a high intensity red light of wavelength 6328 Å ?
Sol. The energy in eV of incident photons is
= 5.47 eV
As stopping potential for ejected electrons is 1.3 V, the maximum kinetic energy of ejected electrons will be
KE_{max} = eV_{0} = 1.3 eV
Now from photoelectric effect equation, we have
E = φ + KE_{max}
or φ = E  KE_{max}
or φ = 5.47  1.3 eV = 4.17 eV
Energy in eV for photons for red light of wavelength 6328 Å is
= 1.96 eV
As E < φ, photocell will not work if irradiated by this red light no matter however intense the light will be.
2. FORCE DUE TO RADIATION (PHOTON)
Each photon has a definite energy and a definite linear momentum. All photons of light of a particular wavelength l have the same energy E = and the same momentum p =
When light of intensity I falls on a surface, it exerts force on that surface. Assume absorption and reflection coefficient of surface be 'a' and 'r' and assuming no transmission.
Assume light beam falls on surface of surface area 'A' perpendicularly as shown in figure.
For calculating the force exerted by beam on surface, we consider following cases.
Case (I) :
a = 1, r = 0
initial momentum of the photon =
final momentum of photon = 0
change in momentum of photon = (upward)
Δp =
energy incident per unit time = IA
no. of photons incident per unit time =Therefore, total change in momentum per unit time = n ΔP =
force on photons = total change in momentum per unit time (upward)
Therefore, force on plate due to photon(F) (downward)
pressure =
Case : (II)
when r = 1, a = 0
initial momentum of the photon = (downward)
final momentum of photon = (upward)
change in momentum
Therefore, energy incident per unit time = I A
no. of photons incident per unit time =
Therefore, total change in momentum per unit time = n. DP =
force = total change in momentum per unit time
(upward on photons and downward on the plate)
pressure
Case : (III)
When 0 < r < 1 a + r = 1
change in momentum of photon when it is reflected = (upward)
change in momentum of photon when it is absorbed = (upward)
no. of photons incident per unit time =
No. of photons reflected per unit time =
No. of photon absorbed per unit time =
force due to absorbed photon (F_{a}) (downward)
Force due to reflected photon (F_{r}) (downward)
total force = F_{a} + F_{r}
Now pressure P
Ex.13 Calculate force exerted by light beam if light is incident on surface at an angle q as shown in figure. Consider all cases.
Sol. Case  I
When a = 1, r = 0
initial momentum of photon (in downward direction at an angle q with vertical) is h/λ
final momentum of photon = 0
change in momentum (in upward direction at an angle q with vertical) =
energy incident per unit time = I A cos θ
Intensity = power per unit normal area
P = I A cos θ
No. of photons incident per unit time =
total change in momentum per unit time (in upward direction at an angle q with vertical)
Force (F) = total change in momentum per unit time
Pressure = normal force per unit Area
Pressure =
Case II
When r = 1, a = 0
Therefore, change in momentum of one photon
(upward)
No. of photons incident per unit time
Therefore,
∴ total change in momentum per unit time
= (upward)
Therefore,
∴ force on the plate (downward)
Pressure
Case III
when 0 < r < 1, a + r = 1
change in momentum of photon when it is reflected = (downward)
change in momentum of photon when it is absorbed = (in the opposite direction of incident beam)
energy incident per unit time = I A cos θ)
no. of photons incident per unit time =
no. of reflected photon (n_{r}) =
no. of absorbed photon (n_{Q}) =
force on plate due to absorbed photons F_{a} = n_{a}. ΔP_{a}
(at an angle θ with vertical )
force on plate due to reflected photons F_{r} = n_{r} ΔP_{r}
(vertically downward)
now resultant force is given by =
and, pressure
Ex.14 A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity I, find the force exerted by the beam on the sphere.
Sol. Let O be the centre of the sphere and OZ be the line opposite to the incident beam (figure). Consider a radius about OZ to get a making an angle q with OZ. Rotate this radius about OZ to get a circle on the sphere. Change q to q +dq and rotate the radius about OZ to get another circle on the sphere. The part of the sphere between these circles is a ring of area 2pr^{2} sinq dq. Consider a small part DA of this ring at P. Energy of light falling on this part in time Dt is
ΔU = IΔt(ΔA cos θ)
The momentum of this light falling on ΔA is ΔU/c along QP. The light is reflected by the sphere along PR. The change in momentum is
(direction along )
The force on ΔA due to the light falling on it, is
(direction along )
The resultant force on the ring as well as on the sphere is along ZO by symmetry. The component of the force on DA along ZO
The force acting on the ring is dF =
The force on the entire sphere is
Note that integration is done only for the hemisphere that faces the incident beam.
3. Debroglie wavelength of matter wave
A photon of frequency v and wavelength λ has energy.
By Einstein's energy mass relation, E = mc^{2} the equivalent mass m of the photon is given by.
. ..(i)
or or ...(ii)
Here p is the momentum of photon. By analogy deBroglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength λ (called deBroglie wavelength and the wave is called matter wave) given by,
...(iii)
where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation,
and a charge q when accelerated by a potential difference V gains a kinetic energy K = qV. Combining all these relations Eq. (iii), can be written as,
(de=Broglie wavelength)....(iv)
3.1 deBroglie wavelength for an electron
If an electron (charge = e) is accelerated by a potential of V volts, it acquires a kinetic energy,
K = eV
Substituting the value of h, m and q in Eq. (iv), we get a simple formula for calculating deBroglie wavelength of an electron.
3.2 deBroglie wavelength of a gas molecule :
let us consider a gas molecule at absolute temperature T. Kinetic energy of gas molecule is given by
K.E. = kT ; k = Boltzman constant
λ_{gas molecules} =
105 videos425 docs114 tests

1. What is the photoelectric effect? 
2. How does the photoelectric effect support the particle nature of light? 
3. What are the key findings of the experimental study on the photoelectric effect? 
4. How does the photoelectric effect support the waveparticle duality of light? 
5. What are the practical applications of the photoelectric effect? 

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