Contents
Physics Topics such as mechanics, thermodynamics, and electromagnetism are fundamental to many other scientific fields.
What are Essential Physical Assumptions needed to Explain the Characteristics of Photoelectric Effect?
Einstein made the following assumptions to explain photo-electric effect.
Einstein’s postulates:
i) A beam of light is incident on a metal surface as streams of photon particles. Energy of each photon having frequency f is, E = hf (h = Planck’s constant).
ii) Incident photons collide with electrons of metal. The collision may produce either of the two effects:
- photon gets reflected with its full energy hf or
- photon transfers entire energy hf to the electron.
Clearly, Einstein used quantum theory of radiation to explain photoelectric effect.
Entire energy hf of the incident photon, when transferred to an electron of the metal, is spent in two ways:
i) A part of the energy is spent to release the electron from the metal whose minimum value is equal to the work function W0 of the metal. But, due to interaction of positive and negative charges inside a metal, most of the electrons need more energy than W0 for the release.
ii) Rest of the energy, changes to kinetic energy of released electrons. These moving electrons are photoelectrons that can set up photoelectric current. If energy absorbed by the electron to leave from the metal surface be the least i.e., W0, emitted electron attains maximum kinetic energy Emax.
Hence, hf = W0 + Emax
or, Emax = hf – W0 …… (1)
If mass of an electron is m and maximum velocity of a photoelectron is vmax, then Emax = \(\frac{1}{2} m v_{\max }^2\). Hence, from equation (1) we get,
\(\frac{1}{2} m v_{\max }^2\) = hf – W0 …… (2)
Also, if V0 is the stopping potential for the incident light of frequency f, then Emax = eV0 (e = charge of an electron)
Hence, from equation (1),
eV0 = hf – W0 ….. (3)
Equations (1), (2) and (3) are practically the same. Each of these is called Einstein’s photoelectric equation.
In most collisions of photons with electrons, there is no energy transfer and the photons are reflected with their full energy hf. Hence probability of photoelectric emission from metal surface is low and the strength of photoelectric current never becomes very high.
Explanation of Photoelectric Effect by Quantum Theory
Einstien’s photoelectric equation is based on quantum theory of radiation. This equation correctly explains the following observations in photoelectric effect.
i) Maximum kinetic energy of photoelectrons: Work function W0 is a constant for a fixed material surface; also frequency of monochromatic light f is a constant. Hence, Emax = hf – W0, is also a constant. Thus, for fixed wave length or fixed frequency of incident light, whatever may be the intensity, emitted photoelectrons cannot attain kinetic energy more than Emax.
ii) Threshold frequency: Work function, W0 is also a constant for a fixed material surface. If frequency of incident light is decreased then as evident from equation Emax = hf – W0, the value of Emax will come down to zero for a certain value of f = f0 (say).
∴ 0 = hf0 – W0 or, hf0 = W0
or, f0 = \(\frac{W_0}{h}\) …… (1)
If value off happens to be below f0, energy of photolectron turns out to be negative and there is no photoelectron emission. Hence, f0 is the threshold frequency. Putting W0 = hf0 in Einstien’s photoelectric equation,
Emax = hf – W0, we get,
Emax = hf – hf0 = h(f – f0) …… (2)
Also, if λ and λ0 be the wavelength of the incident light and threshold wavelength for the metal surface respectively, then
f = \(\frac{c}{\lambda}\) and f0 = \(\frac{c}{\lambda_0}\) [c = speed of light]
Putting these values in equation (2), we get
Emax = hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) ….. (3)
Equations (2) and (3) are the other forms are of Einstein’s photoelectric equation.
iii) Photoelectric emission is instantaneous: Energy transfer takes place between photon of energy hf and electron in the metal due to their elastic collision. Hence, there is no delay in photoelectron emission after the incidence of light.
iv) Dependence of photoelectric current on intensity of incident light: Increase in intensity of incident Light of a constant frequency increases the number of photons incident on the surface of the material. Hence, the number of collisions between the photons and electrons increases. So, more electrons are emitted which increases photoelectric current. This agrees with the results obtained experimentally.
Graphical representations of Einstein’s equation:
i) Frequency (f) versus stopping potential (V0) graph:
From equation (3) in section 1.3.1, we have,
eV0 = hf – W0 or, V0 = \(\frac{h}{e} f\) – \(\frac{W_0}{e}\)
The graph obtained by plotting V0 against f is a straight line of the type, y = mx + c [Fig.]. Knowing the charge of an electron e, Planck’s constant h can be calculated from the slope of the graph \(\frac{h}{e}\). Work function W0 can be obtained from the intercept –\(\frac{W_0}{e}\) in y -axis. Also the intercept with the x -axis gives the threshold frequency f0.
It is important to note that the gradient of the straight line is \(\frac{h}{e}\) for all substances but intercepts from y – and x-axes, –\(\frac{W_0}{e}\) and f0 respectively are different for different substances.
ii) Frequency (f) versus maximum kinetic energy (Emax) graph: From equation (1) in section 1.3.1, we have,
Emax = hf – W0
The graph obtained by plotting Emax against f is a straight line[Fig.]. Comparing the above relation with y = mx + c, we note that slope of the graph is h, x-axis intercept is f0 and y-axis intercept is -W0
Solved Examples
Example 1.
Work function for zinc is 3.6 eV. If threshold frequency for zinc is 9 × 1014 cps, determine the value of Planck’s constant. (1eV = 1.6 × 10-12 erg).
Solution:
Work function, W0 = 3.6eV = 3.6 × 1.6 × 10-12 erg
Threshold frequency, f0 = 9 × 1014 cps = 9 × 1014 Hz
As W0 = hf0,
so, h = \(\frac{W_0}{f_0}\) = \(\frac{3.6 \times 1.6 \times 10^{-12}}{9 \times 10^{14}}\) = 6.4 × 10-27 erg ᐧ s
Example 2.
Maximum kinetic energy of the released photoelectrons emitted from metallic sodium, when a light is incident on it, is 0.73 eV. If the work function of sodium is 1.82 eV, find the energy of the incident photon in eV. Find wavelength of incident light. (h = 6.63 × 10-27 erg.s, 1 eV = 1.6 × 10-12erg) [HS’01]
Solution:
From Einstein’s photoelectric equation, Emax = hf – W0, we get the energy of incident photon as,
E = hf = Emax + W0 = 0.73 + 1.82 = 2.55 eV
Hence wavelength of incident light, λ = \(\frac{c}{f}\) = \(\frac{c}{E / h}\)
∴ λ = \(\frac{h c}{E}\) = \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{2.55 \times 1.6 \times 10^{-12}}\)cm
= \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10} \times 10^8}{2.55 \times 1.6 \times 10^{-12}}\)Å = 4875 Å
Example 3.
Light of wavelength 6000Å is incident on a metal. To release an electron from the metal surface, 1.77 eV of energy is needed. Find the kinetic energy of the fastest photoelectron. What is the threshold frequency of the metal?(h = 6.62 × 10-27 erg ᐧ s, 1eV = 1.6 × 10-12 erg)
Solutton:
Energy of a photon,
hf = h\(\frac{c}{\lambda}\) = \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{6000 \times 10^{-8}}\)erg
= \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{6000 \times 10^{-8} \times 1.6 \times 10^{-12}}\)eV = 2.07eV
As per Einstein’s photoelectric equation,
Emax = hf – W0 = 2.07 – 1.77 = 0.3eV
Threshold frequency,
fo = \(\frac{W_0}{h}\) = \(\frac{1.77 \times 1.6 \times 10^{-12}}{6.62 \times 10^{-27}}\) = 4.28 × 1014 Hz
Example 4.
Photoelectric threshold wavelength for a metal is 3800Å. Find the maximum kinetic energy of emitted photoelectron, when ultraviolet radiation of wave length 2000A is inciden on the metal surface. Planck’s constant, h = 6.62 × 10-34 J ᐧ s. [HS ‘05]
Solution:
Maximum kinetic energy of photoelectron,
Emax = hf – Wo = hf – hf0 = \(\frac{h c}{\lambda}\) – \(\frac{h c}{\lambda_0}\)
= hc\(\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)\) = hc\(\frac{\lambda_0-\lambda}{\lambda \lambda_0}\)
In this case, wavelength of incident light, ‘
λ = 2000 Å = 2000 × 10-10m = 2 × 10-7 m
Threshold wavelength,
λ0 = 3800 × 10-10 m = 3.8 × 10-7m
Hence, Emax = (6.62 × 10-34) × (3 × 108) × \(\frac{(3.8-2) \times 10^{-7}}{3.8 \times 2 \times 10^{-14}}\)J
= \(\frac{6.62 \times 10^{-34} \times 3 \times 10^8 \times 1.8}{3.8 \times 2 \times 10^{-7} \times 1.6 \times 10^{-19}}\)
= 2.94 eV
Example 5.
Threshold wavelength for photoelectric emission from a metal surface is 3800 Å. Ultraviolet light of wave length 2600Å is incident on the metal surface.
(i) Find the work function of the metal and
(ii) maximum kinetic energy of emitted photoelectron. (h = 6.63 × 10-27 erg ᐧ s)
Solution:
i) Threshold wavelength,
λ0 = 3800 Å = 3800 × 10-8 cm
∴ Work function,
W0 = hf0 = h\(\frac{c}{\lambda_0}\) = \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{3800 \times 10^{-8}}\) erg
= \(\frac{6.63 \times 10^{-27} \times 3 \times 10^{10}}{3800 \times 10^{-8} \times 1.6 \times 10^{-12}}\)eV = 3.27 eV
ii) As per Einstein’s photoelectric equation, maximum kinetic energy of photoelectron, Emax = hf – W0
hf = kinetic energy of incident photon
= hf0 × \(\frac{f}{f_0}\) = \(h f_0 \frac{c / \lambda}{c / \lambda_0}\) = \(h f_0 \frac{\lambda_0}{\lambda}\)
= 3.27 × \(\frac{3800}{2600}\) = 4.78 eV
∴ Emax = 4.78 – 3.27 = 1.51 eV
Example 6.
When radiation of wavelength 4940 Å is incident on a metal surface photoelectricity is generated. For a potential difference 0.6V between cathode and anode, photocurrent stops. For another incident radiation, the stopping potential changes to 1.1V. Find work function of the metal and wavelength of the second radiation. (h = 6.6 × 10-27 erg ᐧ s, e = 1.6 × 10-19 C)
Solution:
For the first radiation, stopping potential V0 = 0.6V
∴ Maxinum kinetic energy of photoelectron,
Emax = eV0 = 0.6 eV
Wavelength, λ = 4940 Å = 4940 × 10-8 cm
∴ Energy of incident photon
= hf = h\(\frac{c}{\lambda}\) = \(\frac{6.6 \times 10^{-27} \times 3 \times 10^{10}}{4940 \times 10^{-8}}\)
= \(\frac{6.6 \times 10^{-27} \times 3 \times 10^{10}}{4940 \times 10^{-8} \times 1.6 \times 10^{-12}}\)eV = 2.5 eV
If work function of the metal is W0, from Einstein’s equation,
Emax = hf – W0
or, W0 = hf – Emax = 2.5 – 0.6 = 1.9 eV
For the second radiation, V’0 = 1.1V
Hence, E’max = 1.1 eV
∴ E’max = hf’ – W0
or, hf’ = E’max + W0 = 1.1 + 1.9 = 3.0 eV
Hence, \(\frac{h f}{h f^{\prime}}\) = \(\frac{2.5}{3.0}\) or, \(\frac{f}{f^{\prime}}\) = \(\frac{5}{6}\)
or, \(\frac{c / \lambda}{c / \lambda^{\prime}}\) = \(\frac{5}{6}\) or, \(\frac{\lambda^{\prime}}{\lambda}\) = \(\frac{5}{6}\)
or, λ’ = λ × \(\frac{5}{6}\) = 4940 × \(\frac{5}{6}\) = 4117 Å (approx.)
Example 7.
A stream of photons of energy 10.6 eV and Intensity 2.0 w ᐧ m-2 is incident on a platinum surface. Area of the surface is 1.0 × 10-4 m2 and its work function is 5.6 eV. 0.53% of incident photons emit photoelectrons. Find the number of photoelectrons emitted per second and maximum and minimum energies of the emittedphotoelectronsin eV.(1 eV = 1.6 × 10-19J) [IIT 2000]
Solution:
If intensity of incident light is I, energy incident on a surface area A is IA. Hence, number of photons incident per second, n = \(\frac{I A}{h f}\)
If x% of photons help to emit photoelectrons, number of photo electron emitted per second,
N = n × \(\frac{x}{100}\) = \(\frac{I A x}{h f \cdot 100}\)
Given, I = 2.0W ᐧ m-2, A = 1.0 × 10-4m2,
hf = 10.6 eV = 10.6 × 1.6 × 10-19 J and x = 0.53
∴ N = \(\frac{2.0 \times 1.0 \times 10^{-4} \times 0.53}{10.6 \times 1.6 \times 10^{-19} \times 100}\) = 6.25 × 1011
Minimum kinetic energy of emitted photoelectron = 0
Maximum kinetic energy, Emax = hf – W0 = 10.6 – 5.6 = 5 eV
Example 8.
At what temperature would the kinetic energy of a gas molecule be equal to the energy of photon of wave length 6000 Å Given, Boltzmann’s constant, k = 1.38 × 10-23 J ᐧ K-1; Planck’s constant, h = 6.625 × 10-34 J ᐧ s.
Solution:
Let the required temperature be T K. We know, kinetic energy of a gas molecule
= \(\frac{3}{2}\)kT = \(\frac{3}{2}\) × 1.38 × 10-23 × T
Again kinetic energy of photon
= hf = \(\frac{h c}{\lambda}\) = \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\)
Hence, \(\frac{3}{2}\) × 1.38 × 10-23 × T = \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\)
∴ T = \(\frac{2}{3}\) × \(\frac{6.625 \times 10^{-34} \times 3 \times 10^8}{1.38 \times 10^{-23} \times 6000 \times 10^{-10}}\) = 1.6 × 104 K
Example 9.
Ratio of the work functions of two metal surfaces is 1 : 2. If threshold wavelength of photoelectric effect for the 1st metal is 6000 A, what is the corresponding value for the 2nd metal surface?
Solution:
If work function of 1st and 2nd metals be W0 and W0’, respectively then,
\(\frac{W_0}{W_0^{\prime}}\) = \(\frac{h f_0}{h f_0^{\prime}}\) = \(\frac{h c / \lambda_0}{h c / \lambda_0{ }^{\prime}}\) = \(\frac{\lambda_0^{\prime}}{\lambda_0}\)
or, λ’0 = λ0 × \(\frac{W_0}{W_0^{\prime}}\) = 6000 × \(\frac{1}{2}\) = 3000Å
Example 10.
Work function of a metal surface is 2 eV. Maximum kinetic energy of photoelectrons emitted from the surface for incidence of light of wavelength 4140 Å is 1 eV. What is the threshold wavelength of radiation for that surface?
Solution:
From Einstein’s photoelectric equation,
Emax = hf – W0
or, hf = Emax + W0 = 1 + 2 = 3 eV
Now, \(\frac{h f}{W_0}\) = \(\frac{h f}{h f_0}\) = \(\frac{f}{f_0}\) = \(\frac{c / \lambda}{c / \lambda_0}\) = \(\frac{\lambda_0}{\lambda}\)
∴ λ0 = λ\(\frac{h f}{W_0}\) = 4140 × \(\frac{3}{2}\) = 6210Å