Photons | Class 12 NEB Physics | Complete notes | Short questions and numerical problem solutions

Photons

Quantum nature of radiation

Light has wave character as well as a particle character. Depending on the situation, one of the character is dominant. When light is passed through a double slit, it shows interference which can only be understood by wave theory of light. When light of sufficiently low wavelength falls on a metal surface, electrons are ejected. This phenomena is called Photoelectric effect and can be understood only in terms of the particle nature of light.
According to Planck's quantum theory of radiation, energy from the radiating body is emitted in separate packets of energy, and each packet is called quantum of energy. Each quanta carries a definite amount of energy called Photons . The energy carries by each photon is given by, $$E=hf\text{ ... (i)}$$ where,
f = frequency of radiation,
h = Planck's constant (6.625×10^-34 Js)
Equation (i) can also be expressed in terms of wavelength as, $$E=\frac{hc}{\lambda }\text{ ... (ii)}$$ From equation (i) and (ii), it can be seen that the photon having higher frequency and smaller wavelength has higher energy. Similarly, the photon having smaller frequency and longer wavelength has lower energy.
The particles of light have several properties in common with material particles and several other particles which are different from material particle. The particles of light are called Photons .

Properties of Photons

1. A photon always travel at a speed c $\approx$ 3.0 × 10⁸ m/s in a vacuum.
2. Rest mass of a photon is 0.
3. Each photon has a definite linear momentum and definite energy.
4. A photon may collide with a material particle.
5. The total energy and the total momentum remain conserved in such a collision.

Q.1 Consider a parallel beam of light of wavelength 600 nm. Find the energy of photon.
Solution:
Given,
wavelength of light, λ = 600 nm = 600× 10^-7 m
speed of light, c = 3× 10⁸ m/s
Planck's constant, h = 6.62× 10^-34 Js
energy of photon, E = ?
Now,
$$\begin{array}{rl}E& =\frac{hc}{\lambda }\\ & =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{600\times {10}^{-9}}\\ & =3.31\times {10}^{-19}\\ \therefore E& =2.06875eV\end{array}$$ Hence, the energy of the photon is 2.06875 eV. Here, I have used the relation 1 eV = 1.6× 10^-19 J.

Summary of the photoelectric experiment

Note that, I will not discuss photo electric experiment here. But to begin Einstein's photo electric equation, we must have understanding of the photoelectic experiment briefly which I have summarized as following.

1. When light of sufficiently small wavelength falls on a metal surface, the metal emits electrons. The emissions is almost instantaneous.
2. There is a threshold wavelength λ₀ for a given metal such that if the wavelength of light is more than λ₀, no photoelectric effect takes place.
3. The photocurrent may be stopped by applying a negative potentials to the anode with respect to the cathode. The minimum magnitude of the potential needed to stop the photocurrent is called stopping potential . It is proportional to the maximum kinetic energy of the photoelectrons.
4. The stopping potential does not depend on the intensity of the incident light.This means that the kinetic energy of the photoelectrons is independent of intensity of light.
5. The stopping potential depends on the wavelength of the incident light.
6. The photocurrent increases if the intensity of the incident light is increased.

Work function: The maximum energy that must be given to an electron to take it out of the metal is called the work function of the metal . Work function is denoted by the symbol φ and is different for different metals.

Einstein's photoelectric equation

If you don't mind, I would like to take a tour on the history of Einstein's photoelectric equation which awarded Einstein the Nobel prize in physics for 1921.OK?

Photoelectric effect Image source:Wikipedia
The discovery of photoelectric effect dates back to 1887 when the German physicist Heinrich Rudolf Hertz was working with radio waves. He observed that when UV light shines on two metal electrodes with a voltage applied across them, the light changes the voltage at which sparking takes place. Later in 1902, another German physicist, Philip Lenard demonstrated that electrons are liberated from a metal surface when it is illuminated. Classical theory, which suggests light as an electromagnetic wave was unable to explain this result. Contradictory to the expectation of wave theory, the maximum kinetic energy of liberated electrons was not observed to vary with the intensity of light but was proportional to the frequency of light. These unexpected behaviours led genius mind, Albert Einstein - then a patent clerk in switzerland, to formulate in 1905 a new corpuscular theory of light in which each particle of light, ot photon, contains a fixed amount of energy, or quantum, that depends on the light's frequency. [1]
Five years earlier, Max planck had solved the problem of blackbody radiation by showing that each atom making up the walls of the cavity could only absorb or emit radiation in discrete "quanta" such that the energy of each quantum is given by, $$E=nhf$$ where n is an integer, h is the Planck's constant and f is the frequency of radiation. Planck had thought this relation as a mathematical trick to get theory match with the experiment. But, to everyone surprise, Einstein extended Planck's quanta to light itself. Planck assumed that just the vibrations of the atoms were quantized. In his paper, in March 1905, Einstein said light as a beam of particles whose energies are related to their frequencies according to Planck's formula. When a beam of light is directed at a metal, the photons collides with the atoms. If a photon's frequency is sufficient to knock off an electron, the collision produces the photoelectric effect. As a particle, light carries energy proportional to the frequency of the wave; as a wave it has a frequency determined by the particle's energy . Einstein, for his work on photoelectric effect, was awarded Nobel Prize in physics for 1921. [2]
Citations
[1] Britannica, T. Editors of Encyclopaedia (2020, July 17). Photoelectric effect. Encyclopedia Britannica. https://www.britannica.com/science/photoelectric-effect
[2] Chodos, A., & Ouellette, J. (2005, Jan). 'This Month in Physics History': Einstein and The Photoelectric Effect. APS News . https://www.aps.org/publications/apsnews/200501/history.cfm

Now, lets begin to write the note on this topic.

When a light of frequency f is incident on a metal surface, the photons collide with the free electrons. In a particular collision, the photon may give all of its energy to the free electron. If this energy is greater than the work function φ, the electron may come out of the metal. Here, a part of the energy acquired by the electron is used to pull up the electron from the surface of the metal and rest of it is used in imparting kinetic energy to the emitted electron.
If v_max be the maximum velocity of the photoelectrons, then, $$\begin{array}{rl}E& =\varphi +K.E_{max}\\ hf& =\varphi +K.E_{max}\\ K.Emax& =hf-\varphi \text{ ... (i)}\end{array}$$ Equation (i) is called Einstein's photoelectric equation.
If f₀ be the threshold frequency.
From (i),
$$\begin{array}{rl}K.Emax& =hf-h{f}_0\\ \frac{1}{2}m{v}_{max}^2& =hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})\end{array}$$ Cases:

1. If $\lambda ={\lambda }_0$, $K.E_{max}=0$.
2. If $\lambda >{\lambda }_0$, the energy supplied to the electron is smaller than the work function and no electron will come out.
3. If $\lambda \le {\lambda }_0$, photoelectrons are emitted.

Q.2 Define photoelectric effect. Discuss Einstein's Photoelectir equation. (see the solution above.)

Q.3 Find the maximum wavelength of light that can cause photoelectric effect in sodium. Work function of sodium is 2.3 eV.

Solution:
The maximum wavelength i.e., threshold wavelength is, $$\begin{array}{rl}{\lambda }_0& =\frac{hc}{\varphi }\\ & =\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{2.3\times 1.6\times {10}^{-19}}\\ \therefore {\lambda }_0& =5.396\times {10}^{-7}m\end{array}$$ Hence, the maximum wavelength is 5396 Å.

Millikan's experiment and measurement of Planck's constant

The basic design for the Millikan's experiment for measurement of Planck's constant is shown in figure. It consist of alkali metals placed around a wheel W inside the glass flask. Alkali metals are placed in a vacuum to avoid the formation of oxide films in their surface and a knife K is placed inside the flask to clean the surface of metals.
When monochromatic light falls on the alkali metal, photoelectrons are emitted which reach the electrode C kept negative with respect to the cylindrical blocks. The negative potential of C is increased until the fastest moving electrons stops (repelled back) and current falls to zero. This negative potential is the stopping potential, V₀.
If v_max is the velocity of fastest moving electrons, then the kinetic energy of the fastest photoelectrons is, $$\frac{1}{2}m{v}_{max}^2\text{... (i)}$$ where m is the mass of electron.
Since the stopping potential reduce the kinetic energy of the photoelectrons to zero.
$$\frac{1}{2}m{v}_{max}^2=e{V}_0\text{ ... (ii)}$$ From Einstein's photoelectric equation, $$e{V}_0=hf-h{f}_0\text{ ... (iii)}$$ From this we can write,$$V_0=\frac{h}{e}f-\frac{h}{e}{f}_0\text{ ... (iv)}$$ In equation (iv), $\frac{h}{e}{f}_0$ is a constant for a particular metal.
Comparing equation (iv) with the linear equation $$y=mx+c$$ We get, $$m=\frac{h}{e}$$ and $$c=-\frac{h}{e}{f}_0$$ Also,
$$m=tan\theta$$ So,
$$\begin{array}{rl}tan\theta & =\frac{h}{e}\\ h& =etan\theta \\ \therefore h& =6.625\times {10}^{-34}Js\end{array}$$

Q.4 Write down Einstein's photoelectric equation and describe an experiment to verify it. (See the solution above.)

Q.5 In what ways do photons resemble other particles such as electrons? In what ways do they differ? Do photons have mass? Do they have electric charge? Can they be accelerated? What mechanical properties do they have?
Solution:
Resemblance of photons to other particles:
Photons have momentum and act like particles when they collide with protons and electrons ( remember photoelectric effect ).

Dissimilarity with particles:
Photons are the quanta of light. So, they undergo the phenomena like polarizations while particles do not.

Photons do not have rest mass while they are in motion, they have mass.

No, photons do not have electric charge.

No, photons cannot be accelerated because they always travel at the speed of light (which is constant).

Mechanical properties of photon:
Photons have wavelength, frequency and momentum.

Q.6 According to the photon model, light carries its energy in packets called quanta or photons. Why then we don't see a series of flashes when we look at things?
Solution:
We don't see a series of flashes when we look at things because the number and rate of photons entering our retina are too large to be distinguished distinctly by our brain.

Q.7 During the photoelectric effect, light knocks electrons out of metals. So why don't the metals in your home lose their electrons when you turn on the lights?
Solution:
The energy of a photon is , $$E=hf$$ where, h is the Planck's constant and f is the frquency of light. For the light to knock off electron out of the metals, the photons of light should have very high frequency ( see the relation above! ). The frequency of the light should be larger than the threshold frequency of the metals. This is not the case with the light of our home as it has very low frequency. Hence, the metals in our home do not lose their electrons when we turn on our light.

Q.8 Human skin is relatively insensitive to visible light, but UV radiation can cause severe burns. Does this have anything to do with photon energies? Explain.
Solution:
The frequency of UV radiation is very high and thus is the energy of photon of UV radiation. Due to this reason, the electrons are easily knocked off leaving the atoms in the human skin ionized causing severe burn in the skin.

Q.9 Most black and white photographic film is less sensitive to red light than blue light and has almost no sensitivity to infrared. How can these properties be understood on the basis of photons?
Solution:
The energy of a photon of a light of frequency f is given by, $$E=hf$$ In terms of wavelength, $$E=\frac{hc}{\lambda }$$ The wavelength of blue light is smaller than of red light and thus blue light has higher energy than of red light. Due to this reason, photographic film is more sensitive to blue blue light than red light. Similarly, infrared light has very low energy and hence the film has almost no sensitivity to it.

Q.10 Can a photon be deflected by an electric field? By a magnetic field?
Solution:
Photon is a chargeless particle. Thus, it is not deflected by an electric field or a magnetic field.

Q.11 A hot body is placed in a closed room maintained at a lower temperature. Is the number of photons in the room increasing?
Solution:
When a hot body is placed in a closed room maintained at a lower temperature, heat flows from it to the room. Heat is a form of radiation and the quanta of radiation or light is a photon. Hence, the number of photon increases in this case.

Q.12 It is found that photosynthesis starts in certain plants when exposed to sunlight but it does not start if the plant is exposed only to infrared light. Explain.
Solution:
The photosynthesis initiates when the plant is exposed to the visible light because the photons of visible light has enough energy to stimulate the chlorophyll proteins of the plant leaf. However, the infrared light has very low energy as it has very low frequency as compared to the visible light.

Q.13 If we go on increasing the wavelength of light incident on a metal surface, what changes take place in the number of electrons and energy of the electrons?
Solution:
The energy of a photon of light is, $$E=hf=\frac{hc}{\lambda }$$ When we increase the wavelength of light, the energy of photon decreases. Due to this reason, the energy of the photoelectrons emitted will decrease but the number of photoelectrons emitted will not be affected. However, if the value of wavelength reaches lower than the threshold wavelength, the photoelectrons will not be emitted.

Q.14 What happens to the kinetic energy of the photoelectrons when the intensity of incident light is doubled? Justify your answer.
Solution:
The kinetic energy of the photoelectrons does not depend on the intensity of light. So, there will be no change in the kinetic energy of the photoelectrons when the intensity of light is doubled.

Q.15 Does threshold frequency for photoelectric emission depend on intensity of light? Explain.
Solution:
The work function of a metal is given as, $$\varphi =h{f}_0$$ So we get, $$f_0=\frac{\varphi }{h}$$ where $f_0$ is the threshold frequency. From this relation, we see that the threshold frequency is dependent on the work function of the metal. Work function of the metal is a property of the object / metals. Hence, the threshold frequency does not depend on the intensity of light but on the property of an object, i.e., its work function.

Q.16 In a photoelectric experiment, it was found that the stopping potential decreases from 1.85 V to 0.82 V as the wavelength of the incident light is varied from 300 nm to 400 nm. Calculate the value of the Planck constant from these data.

Solution:
Given,
stopping potential, $V_0$ = 1.85 V at wavelength, $\lambda$ = 300 nm = 300$\times$ 10${}^{-9}$ m
stopping potential, $V_0^{{}^{\prime }}$ = 0.82 V at wavelength, ${\lambda }^{\prime }$ = 400 nm = 400$\times$ 10${}^{-9}$ m
Planck constant, $h$ = ?

From Einstein's photoelectric equation,
$$\begin{array}{rl}E& =\varphi +e{V}_0\\ \frac{hc}{\lambda }& =\varphi +e{V}_0\\ \varphi & =\frac{hc}{\lambda }-e{V}_0\end{array}$$ Again,
$$\begin{array}{rl}{E}^{\prime }& =\varphi +e{V}_0^{\prime }\\ \frac{hc}{{\lambda }^{\prime }}& =\varphi +e{V}_0^{\prime }\\ \varphi & =\frac{hc}{{\lambda }^{\prime }}-e{V}_0^{\prime }\end{array}$$ From this two equations,
$$\begin{array}{rl}\frac{hc}{\lambda }-e{V}_0& =\frac{hc}{{\lambda }^{\prime }}-e{V}_0^{\prime }\\ hc(\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }})& =e(V_0-V_0^{\prime })\\ hc(\frac{1}{300}-\frac{1}{400})\times {10}^9& =1.6\times {10}^{-19}\times (1.85-0.82)\\ h& =\frac{1.6\times {10}^{-19}\times 1.03}{3\times {10}^8\times 8.33\times {10}^5}\\ \therefore h& =6.59\times {10}^{-34}Js\end{array}$$ Hence, the value of the Planck constant estimated from the data is 6.59$\times$10${}^{-34}$ Js.

Q.17 Light of wavelength 4.0×10^-7 m falls on a sodium surface. What is the maximum energy of the emitted electrons in electron volts. (Work function of sodium = 2.3 eV, h = 6.62×10^-34 Js)

Solution:
Given,
wavelength, $\lambda$ = 4.0×10^-7 m
work function, $\varphi$ = 2.3 eV = 2.3$\times$1.6$\times$ 10${}^{-19}$ J
h = 6.62×10^-34 Js
maximum energy, $K_{max}$ = ?

From Einstein's photoelectric equation,
$$\begin{array}{rl}E& =K_{max}+\varphi \\ \frac{hc}{\lambda }& =K_{max}+2.3\times 1.6\times {10}^{-19}\\ \frac{6.62\times {10}^{-34}\times 3\times {10}^8}{4.0\times {10}^{-7}}-3.68\times {10}^{-19}& =K_{max}\\ \therefore K_{max}& =1.28\times {10}^{-19}J\end{array}$$ Hence, the maximum kinetic energy of the emitted electron is 1.28×10^-19 J.
In eV, $$\begin{array}{rl}1eV& =1.6\times {10}^{-19}J\\ 1J& =\frac{1}{1.6\times {10}^{-19}}eV\\ 1.28\times {10}^{-19}& =\frac{1.28\times {10}^{-19}}{1.6\times {10}^{-19}}\\ \therefore K_{max}& =0.8eV\end{array}$$ Q.18 Light of frequency 5.0×10¹⁴ Hz liberates electrons with energy 2.31×10^-19 J from a certain metallic surface. What is the wavelength of UV light which liberates electrons of energy 8.93×10^-19 J from the same surface.

Solution:

frequency, f = 5.0×10¹⁴ Hz
energy of electrons at this frequency, K_max = 2.31×10^-19 J
wavelength, λ' = ? energy of electrons at this wavelength, K_max' = 8.93×10^-19 J

Hints! Use Einstein's photoelectric equation and equate $\varphi$ on both sides.

Q.19 The photoelectric work function of potassium is 2 eV and the surface is illuminated with radiation of wavelength 350 nm. What p.d. have to be applied between a potassium surface and the collecting electrode in order just to prevent collection of electron? What would be the kinetic energy of the electrons?

Solution:
Given,
work function, $\varphi$ = 2 eV = 2 × 1.6 × 10^-19 J
wavelength, $\lambda$ = 350 nm = 350× 10^-9 m >
stopping potential, $V_0$ = ?
kinetic energy of the electrons, $K_{max}$ = ?

From Einstein's photoelectric equation,
$$\begin{array}{rl}E& =e{V}_0+\varphi \\ \frac{hc}{\lambda }& =e{V}_0+\varphi \\ V_0& =\frac{\frac{hc}{\lambda }-\varphi }{e}\\ & =\frac{\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{350\times {10}^{-9}}-3.2\times {10}^{-19}}{1.6\times {10}^{-19}}\\ \therefore V_0& =1.55V\end{array}$$ Hence, the stopping potential is 1.55 V.
Now, the kinetic energy of the electrons is given as, $$K_{max}=e{V}_0$$ Then, $$\begin{array}{rl}{K}_{max}& =e{V}_0\\ & =1.6\times {10}^{-19}\times 1.55\\ \therefore K_{max}& =2.48\times {10}^{-19}J\end{array}$$ Hence, the kinetic energy of the electron is 2.48×10^-19 J.

Q.20 Calculate the energy in electron volts of a quantum of X-radiation of wavelength 0.15 nm. [Take e = 1.6×10^-19 C, h=6.5×10^-34 Js]
Solution:
Given,
wavelength, λ = 0.15 nm = 0.15× 10^-9 m
e = 1.6×10^-19 C
h=6.5×10^-34 Js
energy in electron volts = ?

We have, $$\begin{array}{rl}E& =hf\\ & =\frac{hc}{\lambda }\\ & =\frac{6.5\times {10}^{-34}\times 3\times {10}^8}{0.15\times {10}^{-9}}\\ & =1.3\times {10}^{-15}\end{array}$$ In electron Volts, we can write as, $$\begin{array}{rl}& =\frac{1.3\times {10}^{-15}}{1.6\times {10}^{-19}}\\ & =8125eV\end{array}$$ Hence, the energy of X-radiation in electron volts is 8125 eV.