Text Book Back Questions and Answers - Chapter 7 Dual Nature of Radiation and Matter 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Dual Nature of Radiation and Matter
MultipleChoice Questions
Question 1.
The wavelength $\lambda_e$ of an electron and $\lambda_p$ of a photon of same energy $E$ are related by
(a) $\lambda_{\mathrm{p}} \propto \lambda_{\mathrm{e}}$
(b) $\lambda_{\mathrm{p}} \propto \sqrt{\lambda_e}$
(c) $\lambda_{\mathrm{p}} \propto \frac{1}{\sqrt{\lambda_e}}$
(d) $\lambda_{\mathrm{p}} \propto \lambda_e^2$
Answer:
(d) $\lambda_{\mathrm{p}} \propto \lambda_e^2$
Hint:
de broglie wavelength of electron, $\lambda_{\mathrm{e}}=\frac{h}{\sqrt{2 m E}}$
$
\therefore \text { ie } \lambda_{\mathrm{e}} \propto \frac{1}{\sqrt{E}} \Rightarrow \lambda_e^2 \propto \frac{1}{E} \cdots \cdots
$
de-Broglie wavelength of proton
$
\begin{aligned}
& \lambda_{\mathrm{p}}=\frac{h c}{E} \\
& \lambda_{\mathrm{p}} \propto \frac{1}{E}
\end{aligned}
$
From (1) and (2)
$
\lambda_e^2 \propto \lambda_{\mathrm{p}} \text { i.e., } \lambda_{\mathrm{p}} \propto \lambda_e^2
$

Question 2.
In an electron microscope, the electrons are accelerated by a voltage of $14 \mathrm{kV}$. If the voltage is changed to $224 \mathrm{kV}$, then the de Broglie wavelength associated with the electrons would .......
(a) increase by 2 times
(b) decrease by 2 times
(c) decrease by 4 times
(d) increase by 4 times
Answer:
(c) decrease by 4 times
Hint:
At Voltage, $\mathrm{V}=14 \mathrm{kV}$
de-Broglie wavelength of electron,
$\lambda=\frac{12.3}{\sqrt{14000}} \AA=0.104 \AA$
At voltage, $\mathrm{V}=224 \mathrm{kV}$
$
\begin{aligned}
& \lambda^{\prime}=\frac{12.3}{\sqrt{224000}} \AA=0.026 \AA \\
& \frac{\lambda}{\lambda^{\prime}}=\frac{0.104}{0.0260}=4 \Rightarrow \lambda=4 \lambda^{\prime} \\
& \lambda^{\prime}=\frac{\lambda}{4}
\end{aligned}
$
de-Broglie wavelength of electron is decreased by 4 times
Question 3.
A particle of mass $3 \times 10^{-6} \mathrm{~g}$ has the same wavelength as an electron moving with a velocity

$6 \times 10^6 \mathrm{~ms}^{-1}$ The velocity of the particle is
(a) $1.82 \times 10^{-18} \mathrm{~ms}^{-1}$
(b) $9 \times 10^{-2} \mathrm{~ms}^{-1}$
(c) $3 \times 10^{-31} \mathrm{~ms}^{-1}$
(d) $1.82 \times 10^{-15} \mathrm{~ms}^{-1}$
Answer:
(d) $1.82 \times 10^{-15} \mathrm{~ms}^{-1}$
Hint:
$\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 6 \times 10^6}=1.214 \times 10^{-10} \mathrm{~m}$
Velocity of the particle
$
v=\frac{h}{m \lambda}=\frac{6.63 \times 10^{-34}}{3 \times 10^{-9} \times 1.214 \times 10^{-10}}=1.8204 \times 10^{-15} \mathrm{~ms}^{-1}
$
Question 4.
When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2 \lambda$, the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is
(a) $4 \lambda$
(b) $5 \lambda$
(c) $\frac{5}{2} \lambda$
(d) $3 \lambda$
Answer:
(d) $3 \lambda$
Hint:

$
\begin{aligned}
& e \mathrm{~V}=\frac{h c}{\lambda}-\frac{h c}{\lambda_0} \\
& e\left(\frac{\mathrm{V}}{4}\right)=\frac{h c}{2 \lambda}-\frac{h c}{\lambda_0}
\end{aligned}
$
...(1) $\quad\left[\mathrm{K}_{\max }=h v-\phi_0\right]$
$(1) \div(2)$
$
\frac{4(\mathrm{eV})}{\mathrm{eV}}=\frac{\frac{1}{\lambda}-\frac{1}{\lambda_0}}{\frac{1}{2 \lambda}-\frac{1}{\lambda_0}}=\frac{\left(\frac{\lambda_0-\lambda}{\lambda \lambda_0}\right)}{\left(\frac{\lambda_0-2 \lambda}{2 \lambda \lambda_0}\right)}
$
On solving we get, $\lambda_0=3 \lambda$
Question 5 .
If a light of wavelength $330 \mathrm{~nm}$ is incident on a metal with work function $3.55 \mathrm{eV}$, the electrons are emitted. Then the wavelength of the emitted electron is (Take $h=6.6 \times 10^{-}$ $\left.{ }^{34} \mathrm{Js}\right)$
(a) $<2.75 \times 10^{-9} \mathrm{~m}$
(b) $\geq 2.75 \times 10^{-9} \mathrm{~m}$
(c) $<2.75 \times 10^{-12} \mathrm{um}$
(d) $\leq 2.75 \times 10^{-10} \mathrm{um}$
Answer:
(a) $<2.75 \times 10^{-9} \mathrm{~m}$
Hint:
Maximum $\mathrm{KE}$ of emitted electron is
$
\begin{gathered}
\mathrm{K}_{\max }=\frac{h c}{\lambda}-\phi_0=\left(\frac{1240}{330}-3.55\right) e \mathrm{~V}=(3.76-3.55) e \mathrm{~V} \\
\mathrm{~K}_{\max }=0.21 \mathrm{eV}
\end{gathered}
$
de-Broglie wavelength of emitted electron

$
\begin{aligned}
\lambda=\frac{h}{\sqrt{2 m \mathrm{KE}}} & =\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 0.21 \times 1.6 \times 10^{-19}}}=2.668 \times 10^{-9} \mathrm{~m} \\
\lambda & =2.67 \times 10^{-9} \mathrm{~m}
\end{aligned}
$
The Two wavelength of the emitted electron is $<2.75 \times 10^{-9} \mathrm{~m}$
Question 6.
A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function at the surface of material is
(a) $\frac{h c}{\lambda}$
(b) $\frac{2 h c}{\lambda}$
(c) $\frac{h c}{3 \lambda}$
(d) $\frac{h c}{2 \lambda}$
Answer:
(d) $\frac{h c}{2 \lambda}$
Hint:
$
\begin{aligned}
& \mathrm{KE}_1=\frac{h c}{\lambda}-\Phi \ldots \ldots .(2) \\
& 3 \mathrm{KE}_1=\frac{2 h c}{\lambda}-\Phi \\
& \mathrm{KE}_1=\frac{2 h c}{3 \lambda}-\frac{\Phi}{3 \lambda} \ldots . \text { (2) }
\end{aligned}
$
Equating (1) and (2)
$
\begin{aligned}
& \frac{h c}{\lambda}-\Phi=\frac{2 h c}{3 \lambda}-\frac{\Phi}{3 \lambda} \\
& \frac{h c}{3 \lambda}=\frac{2 \Phi}{3 \lambda} \Rightarrow \Phi=\frac{h c}{2 \lambda}
\end{aligned}
$

Question 7.
In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Th en the maximum possible velocity of the emitted electron will be
(a) $\sqrt{\frac{h v_0}{m}}$
(b) $\sqrt{\frac{6 h v_o}{m}}$
(c) $2 \sqrt{\frac{h v_0}{m}}$
(d) $\sqrt{\frac{h v_0}{2 m}}$
Answer:
(b) $\sqrt{\frac{6 h v_o}{m}}$
Hint:
From Einstein's photoelectric equation
$
\begin{aligned}
\mathrm{K}_{\max } & =h \mathrm{v}-h \mathrm{v}_0 \quad\left[\mathrm{v}=4 \mathrm{v}_0\right] \\
\frac{1}{2} m \mathrm{~V}_{\max }^2 & =4 h \mathrm{v}_0-h \mathrm{v}_0 \\
\mathrm{~V}_{\max }^2 & =\frac{6 h \mathrm{v}_0}{m} \\
\mathrm{~V}_{\max } & =\sqrt{\frac{6 h \mathrm{v}_0}{m}}
\end{aligned}
$

Question 8.
Two radiations with photon energies $0.9 \mathrm{eV}$ and $3.3 \mathrm{eV}$ respectively are falling on a metallic surface successively. If the work function of the metal is $0.6 \mathrm{eV}$, then the ratio of maximum speeds of emitted electrons will be
(a) $1: 4$
(b) $1: 3$
(c) $1: 1$
(d) $1: 9$
Answer:
(b) $1: 3$
Hint:
$
\begin{aligned}
& v_{1_{\max }}=\sqrt{\frac{2(h v-\phi)}{m}}=\sqrt{\frac{2(0.9-0.6)}{m}}=\sqrt{\frac{2(0.3)}{m}} \\
& v_{2_{\max }}=\sqrt{\frac{2(h v-\phi)}{m}}=\sqrt{\frac{2(3.3-0.6)}{m}}=\sqrt{\frac{2(2.7)}{m}}
\end{aligned}
$
$
\begin{aligned}
& \frac{v_{1_{\max }}}{v_{2_{\max }}}=\sqrt{\frac{2(0.3)}{m}} \times \sqrt{\frac{m}{2(2.7)}}=\sqrt{\frac{0.3}{2.7}}=\frac{1}{3} \\
& \left(v_1: v_2\right)_{\max }=1: 3
\end{aligned}
$
Question 9.
A light source of wavelength $520 \mathrm{~nm}$ emits $1.04 \times 10^{15}$ photons per second while the second source of $460 \mathrm{~nm}$ produces $1.38 \times 10^{15}$ photons per second. Then the ratio of power of second source to that of first source is ........
(a) 1.00

(b) 1.02
(c) 1.5
(d) 0.98
Answer:
(c) 1.5
Hint:
Power:
$
\begin{aligned}
& \mathrm{P}=\frac{\mathrm{E}}{t}=\frac{n h \mathrm{v}}{t}=\frac{n}{t} \times \frac{h c}{\lambda} \\
& \mathrm{P}_1=\frac{1.04 \times 10^{15} \times 1240}{520}=2.48 \times 10^{15} \\
& \mathrm{P}_2=\frac{1.38 \times 10^{15} \times 1240}{460}=3.72 \times 10^{15} \\
& \frac{\mathrm{P}_2}{\mathrm{P}_1}=\frac{3.72 \times 10^{15}}{2.48 \times 10^{15}}=1.5
\end{aligned}
$
Question 10.
The mean wavelength of light from sun is taken to be $550 \mathrm{~nm}$ and its mean power is $3.8 \mathrm{x}$ $10^{26} \mathrm{~W}$. The number of photons received by the human eye per second on the average from sunlight is of the order of
(a) $10^{45}$
(b) $10^{42}$
(c) $10^{54}$
(d) $10^{51}$
Answer:
(a) $10^{45}$

$
\begin{aligned}
& \mathrm{P}=\frac{n}{t} \frac{h c}{\lambda} \\
& \therefore \frac{n}{t}=\frac{\mathrm{P} \lambda}{h c}=\frac{3.8 \times 10^{26} \times 550 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8}=1 \times 10^{45}
\end{aligned}
$
Question 11.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 $\mathrm{eV}$ is
(a) $4125 \AA$
(b) $3750 \AA$
(c) $6000 \AA$
(d) $2062.5 \AA$
Answer:
(b) $3750 \AA$
Hint:
$
\begin{aligned}
\phi=h v_0 \Rightarrow \phi & =\frac{h c}{\lambda_0} \\
\therefore \lambda_0=\frac{h c}{\phi} & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3.313 \times 1.6 \times 10^{-19}}=3.7352 \times 10^{-7} \\
& =3.740 \times 10^{-7} \Rightarrow \lambda_0 \approx 3740 \AA
\end{aligned}
$

Question 12.
A light of wavelength $500 \mathrm{~nm}$ is incident on a sensitive plate of photoelectric work function $1.235 \mathrm{eV}$. The kinetic energy of the photo electrons emitted is be (Take $\mathrm{h}=6.6 \mathrm{x}$ $\left.10^{-34} \mathrm{Js}\right)$
(a) $0.58 \mathrm{eV}$
(b) $2.48 \mathrm{eV}$
(c) $1.24 \mathrm{eV}$
(d) $1.16 \mathrm{eV}$
Answer:
(c) $1.24 \mathrm{eV}$
Hint:
$
\begin{aligned}
\mathrm{KE}_{\max } & =\frac{h c}{\lambda}-\phi \\
& =\left(\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9} \times 1.6 \times 10^{-19}}\right)-1.235 \\
& =(2.48-1.235) \mathrm{eV}=1.245 \mathrm{eV}
\end{aligned}
$
Question 13.
Photons of wavelength $\lambda$ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius $\mathrm{R}$ by a perpendicular magnetic field having magnitude $\mathrm{B}$. The work function of the metal is

(a) $\frac{h c}{\lambda}-m_e+\frac{e^2 \mathrm{~B}^2 \mathrm{R}^2}{2 m_e}$
(b) $\frac{h c}{\lambda}+2 m_e\left[\frac{e \mathrm{BR}}{2 m_e}\right]^2$
(c) $\frac{h c}{\lambda}-m_e c^2-\frac{e^2 \mathrm{~B}^2 \mathrm{R}^2}{2 m_e}$
(d) $\frac{h c}{\lambda}-2 m_e\left[\frac{e \mathrm{BR}}{2 m_e}\right]^2$
Answer:
(d) $\frac{h c}{\lambda}-2 m_e\left[\frac{e \mathrm{BR}}{2 m_e}\right]^2$
Hint:
Magnetic lorentz force $=$ Centripetal force
$
\begin{aligned}
\mathrm{BqV} & =\frac{m v^2}{\mathrm{R}} \\
\therefore \quad \mathrm{V} & =\frac{\mathrm{qBR}}{m}=\frac{\mathrm{eBR}}{m}
\end{aligned}
$
From Einstein's photo electric equation

$
\begin{aligned}
\mathrm{KE}_{\max } & =\frac{h c}{\lambda}-\phi \\
\phi & =\frac{h c}{\lambda}-\frac{1}{2} m_e \mathrm{~V}^2=\frac{h c}{\lambda}-\frac{1}{2} m_e\left(\frac{e \mathrm{BR}}{m_e}\right)^2 \\
\phi & =\frac{h c}{\lambda}-2 m_e\left(\frac{e \mathrm{BR}}{2 m_e}\right)^2
\end{aligned}
$
Question 14.
The work functions for metals $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5.0 \mathrm{eV}$ respectively.
The metals which will emit photoelectrons for a radiation of wavelength $4100 \AA$ is/are
(a) A only
(b) both $\mathrm{A}$ and $\mathrm{B}$
(c) all these metals
(d) none
Answer:
(b) both $\mathrm{A}$ and $\mathrm{B}$
Hint:
Energy of radiation
$
\mathrm{E}=h \mathrm{v}=\frac{h c}{\lambda}=\left(\frac{\left(6.6 \times 10^{-34} \times 3 \times 10^8\right) / 1.6 \times 10^{-19}}{410 \times 10^{-9}}\right)=\frac{1240}{410}
$
$\mathrm{E}=3.04 \mathrm{eV}$
Since energy of incident radiation is greater than the work function of metals A and B. So metal A and B will emit photoelectrons.

Question 15.
Emission of electrons by the absorption of heat energy is called emission.
(a) photoelectric
(b) field
(c) thermionic
(d) secondary
Answer:
(c) thermionic
Short AnswerQuestions
Question 1.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.
Question 2.
Define work function of a metal. Give its unit.
Answer:
The minimum energy needed for an electron to escape from the metal surface is called work function of that metal. It's unit is electron volt $(\mathrm{eV})$.
Question 3.
What is photoelectric effect?
Answer:
The ejection of electrons from a metal plate when illuminated by light or any other electromagnetic radiation of suitable wavelength (or frequency) is called photoelectric effect.

Question 4.
How does photocurrent vary with the intensity of the incident light?
Answer:
Photocurrent - the number of electrons emitted per second is directly proportional to the intensity of the incident light.
Question 5.
Give the definition of intensity of light and its unit.
Answer:
Intensity of light refer to the strength or brightness or amount of light produced by a specific source. It's unit is candela (cd)
Question 6.
How will you define threshold frequency?
Answer:
For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
Question 7.
What is a photo cell? Mention the different types of photocells.
Answer:
photocells: Photo electric cell or photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect.
Types:
- Photo emissive cell
- Photo voltaic cell
- Photo conductive cell
Question 8.
Write the expression for the de Broglie wavelength associated with a charged particle of charge $\mathrm{q}$ and mass $\mathrm{m}$, when it is accelerated through a potential $\mathrm{V}$.
Answer:
An electron of mass $\mathrm{m}$ is accelerated through a potential difference of $\mathrm{V}$ volt. The kinetic energy acquired by the electron is given by
$
\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}
$

Therefore, the speed $\mathrm{v}$ of the electron is $\mathrm{v}=\sqrt{\frac{2 e v}{m}}$ Hence, the de Broglie wavelength of the electron is $\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m V}}$
Question 9.
State de Broglie hypothesis.
Answer:
De Broglie hypothesis, all matter particles like electrons, protons, neutrons in motion are associated with waves.
Question 10.
Why we do not see the wave properties of a baseball?
Answer:
Due to the large mass of a baseball, the de Broglie wavelength $\left[\lambda=\frac{h}{m v}\right]$ associated with a moving baseball is very small. Hence its wave nature is not visible.
Question 11.
A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.
Answer:
de-Broglie wavelength of the particle is $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K}}$
i.e. $\lambda \propto \frac{h}{\sqrt{m}}$
As $\mathrm{m}_{\mathrm{e}}<<\mathrm{m}_{\mathrm{p}}$, so $\lambda_{\mathrm{e}}>\lambda_{\mathrm{p}}$
Hence protons have greater de-Broglie wavelength.
Question 12.
Write the relationship of de Broglie wavelength $\lambda$ associated with a particle of mass $\mathrm{m}$ in terms of its kinetic energy $\mathrm{K}$.
Answer:
Kinetic energy of the particle, $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2=\frac{P^2}{2 m}$
$\mathrm{p}=\sqrt{2 m K}$
de-Broglie wavelength of the particle $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K}}$

Question 13.
Name an experiment which shows wave nature of the electron. Which phenomenon was observed in this experiment using an electron beam?
Answer:
- Davisson - Germer experiment confirmed the wave nature of electrons.
- They demonstrated that electron beams are diffracted when they fall on crystalline solids.
Question 14.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?
Answer:
$
\left[\lambda=\frac{h}{p}\right]
$
Kinetic energy of the particle $\mathrm{K}=\frac{1}{2} \mathrm{mv}^2=\frac{P^2}{2 m}=\frac{h^2}{2 m \lambda^2}$
i.e. $\lambda=\frac{h}{\sqrt{2 m K}} ; \lambda \propto \frac{1}{\sqrt{m}}$
$
\frac{\lambda_e}{\lambda_\alpha}=\sqrt{\frac{m_\alpha}{m_e}}
$
Long Answer Questions
Question 1.
What do you mean by electron emission? Explain briefly various methods of electron emission.
Answer:
Electron emission:
1. Free electrons possess some kinetic energy and this energy is different for different electrons. The kinetic energy of the free electrons is not sufficient to overcome the surface barrier.

2. Whenever an additional energy is given to the free electrons, they will have sufficient energy to cross the surface barrier. And they escape from the metallic surface.
3. The liberation of electrons from any surface of a substance is called electron emission.
There are mainly four types of electron emission which are given below.
(i) Thermionic emission:
When a metal is heated to a high temperature, the free electrons on the surface of the metal get sufficient energy in the form of thermal energy so that they are emitted from the metallic surface. This type of emission is known as thermionic emission.

The intensity of the thermionic emission (the number of electrons emitted) depends on the metal used and its temperature.

Examples: cathode ray tubes, electron microscopes, $\mathrm{X}$-ray tubes etc.
(ii) Field emission:
Electric field emission occurs when a very strong electric field is applied across the metal. This strong field pulls the free electrons and helps them to overcome the surface barrier of the metal.

Examples: Field emission scanning electron microscopes, Field-emission display etc.

(iii) Photo electric emission:
When an electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is transferred from the radiation to the free electrons. Hence, the free electrons get sufficient energy to cross the surface barrier and the photo electric emission takes place. The number of electrons emitted depends on the intensity of the incident radiation.

Examples: Photo diodes, photo electric cells etc.
(iv) Secondary emission:
When a beam of fast moving electrons strikes the surface of the metal, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface. Thus the free electrons get sufficient kinetic energy so that the secondary emission of, electron occurs.

Examples: Image intensifies, photo multiplier tubes etc.
Question 2.
Briefly discuss the observations of Hertz, Hallwachs and Lenard.
Answer:
Hertz observation:
1. In 1887, Heinrich Hertz first became successful in generating and detecting electromagnetic wave with his high voltage induction coil to cause a spark discharge between two metallic spheres.
2. When a spark is formed, the charges will oscillate back and forth rapidly and the electromagnetic waves are produced.
3. The electromagnetic waves thus produced were detected by a detector that has a copper wire bent in the shape of a circle. Although the detection of waves is successful, there is a problem in observing the tiny spark produced in the detector.
4. In order to improve the visibility of the spark, Hertz made many attempts and finally noticed an important thing that small detector spark became more vigorous when it was exposed to ultraviolet light.
5. The reason for this behaviour of the spark was not known at that time. Later it was found that it is due to the photoelectric emission.
6. Whenever ultraviolet light is incident on the metallic sphere, the electrons on the outer surface are emitted which caused the spark to be more vigorous.
Hallwachs' observation:

8. In 1888 , Wilhelm Hallwachs, a German physicist, confirmed that the strange behaviour of the spark is due to the action of ultraviolet light with his simple experiment.
9. A clean circular plate of zinc is mounted on an insulating stand and is attached to a gold leaf electroscope by a wire. When the uncharged zinc plate is irradiated by ultraviolet light from an arc lamp, it becomes positively charged and the leaves will open.
10. Further, if the negatively charged zinc plate is exposed to ultraviolet light, the leaves will close as the charges leaked away quickly. If the plate is positively charged, it becomes more positive upon UV rays irradiation and the leaves will open further.
11. From these observations, it was concluded that negatively charged electrons were emitted from the zinc plate under the action of ultraviolet light.
Lenard's observation:
1. In 1902, Lenard studied this electron emission phenomenon in detail. The apparatus consists of two metallic plates $\mathrm{A}$ and $\mathrm{C}$ placed in an evacuated quartz bulb. The galvanometer $\mathrm{G}$ and battery B are connected in the circuit.
2. When ultraviolet light is incident on the negative plate $\mathrm{C}$, an electric current flows in the circuit that is indicated by the deflection in the galvanometer. On other hand, if the positive plate is irradiated by the ultraviolet light, no current is observed in the circuit.
3. From these observations, it is concluded that when ultraviolet light falls on the negative plate, electrons are ejected from it which are attracted by the positive plate A. On reaching the positive plate through the evacuated bulb, the circuit is completed and the current flows in it.
4. Thus, the ultraviolet light falling on the negative plate causes the electron emission from the surface of the plate.

Question 3.
Explain the effect of potential difference on photoelectric current.
Answer:
Effect of potential difference on photoelectric current:
1. To study the effect of potential difference $\mathrm{V}$ between the electrodes on photoelectric current, the frequency and intensity of the incident light are kept constant. Initially the potential of $\mathrm{A}$ is kept positive with respect to $\mathrm{C}$ and the cathode is irradiated with the given light.
2. Now, the potential of $\mathrm{A}$ is increased and the corresponding photocurrent is noted. As the potential of $\mathrm{A}$ is increased, photocurrent is also increased. However a stage is reached where photocurrent reaches a saturation value (saturation current) at which all the photoelectrons from $\mathrm{C}$ are collected by $\mathrm{A}$. This is represented by the flat portion of the graph between potential of $\mathrm{A}$ and photocurrent.
3. When a negative (retarding) potential is applied to $\mathrm{A}$ with respect to $\mathrm{C}$, the current does not immediately drop to zero because the photoelectrons are emitted with some definite and different kinetic energies.
4. The kinetic energy of some of the photoelectrons is such that they could overcome the retarding electric field and reach the electrode $A$.
5. When the negative (retarding) potential of $\mathrm{A}$ is gradually increased, the photocurrent starts to decrease because more and more photoelectrons are being repelled away from reaching the electrode $\mathrm{A}$. The photocurrent becomes zero at a particular negative potential V0, called stopping or cut-off potential.

6. Stopping potential is that the value of the negative (retarding) potential given to the collecting electrode A which is just sufficient to stop the most energetic photoelectrons emitted and make the photocurrent zero.
7. At the stopping potential, even the most energetic electron is brought to rest. Therefore, the initial kinetic energy of the fastest electron $\left(\mathrm{K}_{\max }\right)$ is equal to the work done by the stopping potential to stop it $\left(\mathrm{eV}_0\right)$.
$
\mathrm{K}_{\max }=\frac{1}{2} m v_{\max }^2=\mathrm{eV}_0 \ldots \text { (1) }
$
Where $v_{\max }$ is the maximun speed of the emitted photoelectron.
$
\begin{aligned}
& v_{\max }=\sqrt{\frac{2 e \mathrm{~V}_0}{m}} \\
& v_{\max }=\sqrt{\frac{2 \times 1.602 \times 10^{-19}}{9.1 \times 10^{-31}} \times \mathrm{V}_0} \\
& =5.93 \times 10^5 \sqrt{\mathrm{V}_0} \\
& =5.93 \times 10^5 \sqrt{V_0} \ldots . \text { (2) } \\
&
\end{aligned}
$
8. From the graph, when the intensity of the incident light alone is increased, the saturation current also increases but the value of $\mathrm{V}_0$ remains constant.
9. Thus, for a given frequency of the incident light, the stopping potential is independent of intensity of the incident light. This also implies that the maximum kinetic energy of the photoelectrons is independent of intensity of the incident light.

Question 4.
Explain how frequency of incident light varies with stopping potential.
Answer:
Effect of frequency of incident light on stopping potential:
1. To study the effect of frequency of incident light on stopping potential, the intensity of the incident light is kept constant. The variation of photocurrent with the collector electrode potential is studied for radiations of different frequencies and a graph drawn between them. From the graph, it is clear that stopping potential vary over different frequencies of incident light.

2. Greater the frequency of the incident radiation, larger is the corresponding stopping potential. This implies that as the frequency is increased, the photoelectrons are emitted with greater kinetic energies so that the retarding potential needed to stop the photoelectrons is also greater.

3. Now a graph is drawn between frequency and the stopping potential for different metals. From this graph, it is found that stopping potential varies linearly with frequency. Below a certain frequency called threshold frequency, no electrons are emitted; hence stopping potential is zero for that reason. But as the frequency is increased above threshold value, the stopping potential varies linearly with the frequency of incident light.
Question 5 .
List out the laws of photoelectric effect.
Answer:
Laws of photoelectric effect:
1. For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
2. Maximum kinetic energy of the photo electrons is independent of intensity of the incident light.

3. Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
4. For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
5. There is no time lag between incidence of light and ejection of photoelectrons.
Question 6.
Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Answer:
Failures of classical wave theory:
From Maxwell's theory, light is an electromagnetic wave consisting of coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behaviour. Let us try to explain the experimental observations of photoelectric effect using wave picture of light.
1. When light is incident on the target, there is a continuous supply of energy to the electrons. According to wave theory, light of greater intensity should impart greater kinetic energy to the liberated electrons (Here, Intensity of light is the energy delivered per unit area per unit time). But this does not happen. The experiments show that maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.
2. According to wave theory, if a sufficiently intense beam of light is incident on the surface, electrons will be liberated from the surface of the target, however low the frequency of the radiation is. From the experiments, we know that photoelectric emission is not possible below a certain minimum frequency. Therefore, the wave theory fails to explain the existence of threshold frequency.
3. Since the energy of light is spread across the wavefront, the electrons which receive energy from it are large in number. Each electron needs considerable amount of time (a few hours) to get energy sufficient to overcome the work function and to get liberated from the surface. But experiments show that photoelectric emission is almost instantaneous process (the time lag is less than $10 " 9 \mathrm{~s}$ after the surface is illuminated) which could not be explained by wave theory.
Question 7.
Explain the quantum concept of light.
Answer:
Concept of quantization of energy:
Max Planck proposed quantum concept in 1900 in order to explain the thermal radiations emitted by a black body and the shape of its radiation curves. According to Planck, matter is composed of a large number of oscillating particles (atoms) which vibrate with different frequencies. Each atomic oscillator - which vibrates with its characteristic frequency - emits or absorbs electromagnetic radiation of the same frequency. It also says that
1. If an oscillator vibrates with frequency $\mathrm{v}$, its energy can have only certain discrete values, given by the equation.
$
\mathrm{E}_{\mathrm{n}}=\mathrm{nhv} \mathrm{n}=1,2,3 \ldots \ldots \ldots
$
where A is a constant, called Planck's constant.
2. The oscillators emit or absorb energy in small packets or quanta and the energy of each quantum is $\mathrm{E}=\mathrm{hv}$.
This implies that the energy of the oscillator is quantized - that is, energy is not continuous as believed in the wave picture. This is called quantization of energy.
Question 8.
Obtain Einstein's photoelectric equation with necessary explanation. Einstein's explanation of photoelectric equation:
Answer:
1. When a photon of energy hv is incident on a metal surface, it is completely absorbed by a single electron and the electron is ejected.
2. In this process, a part of the photon energy is used for the ejection of the electrons from the metal surface (photoelectric work function $\Phi_0$ ) and the remaining energy as the kinetic energy of the ejected electron. From the law of conservation of energy,
$
\mathrm{hv}=\Phi_0+\frac{1}{2} \mathrm{mv}^2
$

where $m$ is the mass of the electron and $u$ its velocity

3. If we reduce the frequency of the incident light, the speed or kinetic energy of photo electrons is also reduced. At some frequency $V_0$ of incident radiation, the photo electrons are ejected with almost zero kinetic energy. Then the equation (1) becomes $h v_0=\Phi_0$
where vQ is the threshold frequency. By rewriting the equation (1), we get $\mathrm{hv}=\mathrm{h} v_0+\frac{1}{2} \mathrm{mv}^2 \ldots \ldots(2)$
The equation (2) is known as Einstein's Photoelectric equation.
If the electron does not lose energy by internal collisions, then it is emitted with maximum kinetic energy $\mathrm{K}_{\max }$. Then
$
\mathrm{K}_{\max }=\frac{1}{2} m v_{\max }^2
$
where nmax is the maximum velocity of the electron ejected. The equation (1) is rearranged as follows:
$
\mathrm{K}_{\max }=\mathrm{h} v-\Phi_0
$

Question 9.
Explain experimentally observed facts of photoelectric effect with the help of Einstein's explanation.
Answer:
Explanation for the photoelectric effect:
The experimentally observed facts of photoelectric effect can be explained with the help of . Einstein's photoelectric equation.
1. As each incident photon liberates one electron, then the increase of intensity of the light (the number of photons per unit area per unit time) increases the number of electrons emitted thereby increasing the photocurrent. The same has been experimentally observed.
2. From $\mathrm{K}_{\max }=\mathrm{h} v-\Phi_0$, it is evident that $\mathrm{Kmax}$ is proportional to the frequency of the light and is independent of intensity of the light.
3. As given in Einstein's photoelectric equation, there must be minimum energy (equal to the work function of the metal) for incident photons to liberate electrons from the metal surface. Below which, emission of electrons is not possible. Correspondingly, there exists minimum frequency called threshold frequency below which there is no photoelectric emission.
4. According to quantum concept, the transfer of photon energy to the electrons is instantaneous so that there is no time lag between incidence of photons and ejection of electrons.
Question 10.
Give the construction and working of photo emissive cell.
Answer:
Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.
Construction:
1. It consists of an evacuated glass or quartz bulb in which two metallic electrodes - that is, a cathode and an anode are fixed.

2. The cathode $\mathrm{C}$ is semi-cylindrical in shape and is coated with a photo sensitive material. The anode $\mathrm{A}$ is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
3. A potential difference is applied between the anode and the cathode through a galvanometer $\mathrm{G}$.

Working:
4. When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer.
5. For a given cathode, the magnitude of the current depends on
(i) the intensity to incident radiation and (ii) the potential difference between anode and cathode.
Question 11.
Derive an expression for de Broglie wavelength of electrons.
Answer:
An electron of mass $m$ is accelerated through a potential difference of $\mathrm{V}$ volt. The kinetic energy acquired by the electron is given by
$\frac{1}{2} \mathrm{mv}^2=$ evacuated
Therefore, the speed $\mathrm{v}$ of the electron is $\mathrm{v}=\sqrt{\frac{2 e V}{m}}$
Hence, the de Broglie wavelength of the electron is $\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m V}}$
Substituting the known values in the above equation, we get
$
\begin{aligned}
& \lambda=\frac{6.626 \times 10^{-34}}{\sqrt{2 \mathrm{~V} \times 1.6 \times 10^{-19} \times 9.11 \times 10^{-31}}} \\
& \lambda=\frac{12.27 \times 10^{-10}}{\sqrt{\mathrm{V}}} \text { meter (or } \lambda=\frac{12.27}{\sqrt{\mathrm{V}}} \AA
\end{aligned}
$
For example, if an electron is accelerated through a potential difference of $100 \mathrm{~V}$, then its de Broglie wavelength is $1.227 \mathrm{~A}$. Since the kinetic energy of the electron, $\mathrm{K}=\mathrm{eV}$, then the de Broglie wavelength associated with electron can be also written as
$
\lambda=\frac{h}{\sqrt{2 m K}}
$

Question 12 .
Briefly explain the principle and working of electron microscope.
Answer:
Electron Microscope:
Principle:
1. This is the direct application of wave nature of particles. The wave nature of the electron is used in the construction of microscope called electron microscope.
2. The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
3. De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes.
4. As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
5. Electron microscopes giving magnification more than 2,00.000 times are common in research laboratories.

Working:
1. The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
2. The electrons emitted from the source are accelerated by high potentials. The beam is made parallel by magnetic condenser lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
3. With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science.
Question 13.
Describe briefly Davisson - Germer experiment which demonstrated the wave nature of electrons.
Answer:
Davisson - Germer experiment:
1. De Broglie hypothesis of matter waves was experimentally confirmed by Clinton Davisson and Lester Germer in 1927. They demonstrated that electron beams are diffracted when they fall on crystalline solids.
2. Since ciystal can act as a three-dimensional diffraction grating for matter waves, the electron waves incident on crystals are diffracted off in certain specific directions.

3. The filament $\mathrm{F}$ is heated by a low tension (L.T.) battery. Electrons are emitted from the hot filament by thermionic emission. They are then accelerated due to the potential difference between the filament and the anode aluminium cylinder by a high tension (H.T.) battery.
4. Electron beam is collimated by using two thin aluminium diaphragms and is allowed to strike a single crystal of Nickel.
5. The electrons scattered by $\mathrm{Ni}$ atoms in different directions are received by the electron detector which measures the intensity of scattered electron beam.
6. The detector is rotatable in the plane of the paper so that the angle $\Phi$ between the incident beam and the scattered beam can be changed at our will.
7. The intensity of the scattered electron beam is measured as a function of the angle $\theta$.

8. From the graph shows the variation of intensity of the scattered electrons with the angle 0 for the accelerating voltage of $54 \mathrm{~V}$. For a given accelerating voltage $\mathrm{V}$, the scattered wave shows a peak or maximum at an angle of $50^{\circ}$ to the incident electron beam.
This peak in intensity is attributed to the constructive interference of electrons diffracted from various atomic layers of the target material.
9. From the known value of interplanar spacing of Nickel, the wavelength of the electron wave has been experimentally calculated as $1.65 \AA$.
10. The wavelength can also be calculated from de Broglie relation for $\mathrm{V}=54 \mathrm{~V}$ from equation.
$
\begin{aligned}
& \lambda=\frac{12.27}{\sqrt{V}} \AA=\frac{12.27}{\sqrt{54}} \\
& \lambda=1.67 \AA
\end{aligned}
$
11. This value agrees well with the experimentally observed wavelength of $1.65 \AA$. Thus this experiment directly verifies de Broglie's hypothesis of the wave nature of moving particles.