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Cathode Rays and Electron Emission Questions in English

Class 12 Physics · Dual Nature of Radiation and matter · Cathode Rays and Electron Emission

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51
Medium
Write information about the negative electric charge emitted from metals when $UV$ light is incident on them or when metals are heated.

Solution

(N/A) In $1887$,it was observed that when ultraviolet $(UV)$ light is incident on a metal surface or when some metals are heated,negative charges are emitted from them.
Particles produced by these different methods exhibit similar characteristics. In $1897$,$J.J. Thomson$ identified these particles as electrons and demonstrated that the electron is a fundamental and universal constituent of matter. For the discovery of the electron,$J.J. Thomson$ was awarded the Nobel Prize in Physics in $1906$.
In $1913$,the American scientist $R.A. Millikan$ measured the charge of an electron using his oil drop experiment. He observed that the charge on any oil drop is always an integral multiple of the fundamental charge $e = 1.602 \times 10^{-19} \ C$. Thus,$Millikan$ proved that electric charge is quantized.
By knowing the specific charge $\frac{e}{m}$ and the elementary charge $e$,the mass of the electron $(m)$ can be calculated.
52
EasyMCQ
In $1897$,$J$.$J$. Thomson discovered which particle?
A
Proton
B
Electron
C
Neutron
D
Positron

Solution

(B) In $1897$,$J$.$J$. Thomson conducted experiments with cathode ray tubes. By observing the deflection of cathode rays in electric and magnetic fields,he determined that they consisted of negatively charged particles. He named these particles 'electrons'. Thus,$J$.$J$. Thomson is credited with the discovery of the electron.
53
MediumMCQ
Atom or molecules can be visualized by which device or instrument?
A
Scanning Tunneling Microscope
B
Optical Microscope
C
Electron Microscope
D
Telescope

Solution

(A) The Scanning Tunneling Microscope $(STM)$ is a powerful instrument used to image surfaces at the atomic level. It allows scientists to visualize individual atoms and molecules by measuring the tunneling current between a sharp metallic tip and the sample surface. Unlike optical microscopes,which are limited by the wavelength of visible light,the $STM$ provides atomic-scale resolution.
54
EasyMCQ
Which scientist invented cathode rays?
A
$J$.$J$. Thomson
B
William Crookes
C
Ernest Rutherford
D
James Chadwick

Solution

(B) Cathode rays were first observed and investigated by $William$ $Crookes$ in the late $19^{th}$ century. While $J.J.$ $Thomson$ is famous for discovering the electron within these rays,the invention and initial study of the cathode ray tube and the rays themselves are attributed to $William$ $Crookes$.
55
Medium
What is specific charge? Write its standard value.

Solution

(N/A) The specific charge of a particle is defined as the ratio of its charge $(q)$ to its mass $(m)$.
Mathematically,it is expressed as: $\text{Specific charge} = \frac{q}{m}$.
The $SI$ unit of specific charge is $\text{C/kg}$.
For an electron,the specific charge is approximately $1.76 \times 10^{11} \ \text{C/kg}$.
56
EasyMCQ
What is the speed of cathode rays?
A
Equal to the speed of light
B
Approximately $0.1c$ to $0.3c$
C
Zero
D
Greater than the speed of light

Solution

(B) Cathode rays are streams of fast-moving electrons.
Their speed depends on the potential difference applied between the cathode and the anode.
Typically, the speed of cathode rays ranges from $1/10$ to $1/3$ of the speed of light $(c)$, which is approximately $3 \times 10^7 \, m/s$ to $1 \times 10^8 \, m/s$.
57
Medium
Write the name of the universal and fundamental particle of matter.

Solution

(N/A) The universal and fundamental particle of matter is the $electron$.
It is considered a fundamental particle because it is a lepton and cannot be further subdivided into smaller components.
It is universal because it is present in all atoms of all elements.
58
MediumMCQ
How can the mass of an electron be obtained?
A
By measuring the charge of the electron $(e)$ and the specific charge $(e/m)$.
B
By measuring the velocity of the electron.
C
By measuring the kinetic energy of the electron.
D
By measuring the wavelength of the electron.

Solution

(A) The mass of an electron $(m)$ can be determined by using the values of the charge of the electron $(e)$ and the specific charge $(e/m)$.
$1$. The charge of an electron $(e)$ is determined using Millikan's oil drop experiment,which gives $e \approx 1.602 \times 10^{-19} \ C$.
$2$. The specific charge $(e/m)$ is determined using $J$.$J$. Thomson's experiment (cathode ray tube experiment),which gives $e/m \approx 1.758 \times 10^{11} \ C/kg$.
$3$. By dividing the charge by the specific charge,we get the mass: $m = \frac{e}{(e/m)}$.
$4$. Substituting the values: $m = \frac{1.602 \times 10^{-19}}{1.758 \times 10^{11}} \approx 9.11 \times 10^{-31} \ kg$.
59
Medium
Explain the emission of electrons from a metal. Define work function. Write its unit,and state the factors upon which the work function of a metal depends.

Solution

(N/A) Normally,electrons cannot escape from the surface of a metal because they are bound by the attractive forces of the positive ions in the metal lattice.
When the temperature of a metal is increased,the kinetic energy of the electrons due to thermal oscillation increases. When this energy exceeds the binding energy,the electron can escape from the metal surface,leaving the metal positively charged.
However,due to the electrostatic attractive force between the newly formed positive ion and the escaping electron,the electron is typically pulled back to the metal surface.
An electron can only permanently escape from the metal surface if its energy is greater than the potential energy barrier created by these attractive forces.
Work function: The minimum amount of energy required to remove an electron from the surface of a metal is called the work function of that metal.
Work function is denoted by $\phi_{0}$. Its $SI$ unit is $eV$ (electron volt).
Factors affecting work function: The work function of a metal depends on the nature of the metal and the condition of its surface (such as impurities or surface finish).
60
Easy
Write and explain the methods to obtain the emission of electrons from a metal surface.

Solution

(N/A) The emission of electrons from a metal surface can be obtained by the following methods:
$(1)$ Thermionic emission: When metals are heated (e.g.,by passing a current through a metal filament),electrons are provided with sufficient thermal energy to overcome the work function and escape from the metal surface. This process is called thermionic emission.
Example: Electrons are emitted in diodes,triodes,or $TV$ tubes using thermionic emission.
$(2)$ Field emission: When a very strong electric field of the order of $10^{8} \text{ V/m}$ is applied to a metal surface,electrons are pulled out from the surface. This process is called field emission.
$(3)$ Photoelectric emission: When electromagnetic radiation of a suitable frequency (above the threshold frequency) is incident on a clean metal surface,electrons are emitted from it. This process is called photoelectric emission,and the emitted electrons are known as photoelectrons.
61
MediumMCQ
What is the value (magnitude) of the electric field required for field emission?
A
$10^8 \ V/m$
B
$10^6 \ V/m$
C
$10^4 \ V/m$
D
$10^2 \ V/m$

Solution

(A) Field emission is a process where electrons are emitted from the surface of a metal due to the application of a very strong external electric field.
This strong electric field pulls the electrons out of the metal surface by reducing the potential barrier.
The magnitude of the electric field required for this process is typically of the order of $10^8 \ V/m$.
Therefore,the correct option is $A$.
62
Medium
The filament of an electron gun is made of which material and coated with which material? Why?

Solution

(N/A) The filament of an electron gun is typically made of $Tungsten$ $(W)$.
It is coated with $Barium$ $Oxide$ $(BaO)$ or $Strontium$ $Oxide$ $(SrO)$.
Reason: $Tungsten$ has a very high melting point $(3695 \ K)$,which allows it to withstand high temperatures without melting. However,its work function is high,meaning it requires a lot of energy to emit electrons. By coating the filament with $Barium$ $Oxide$ or $Strontium$ $Oxide$,the work function of the surface is significantly reduced. This allows for the emission of a large number of electrons at relatively lower temperatures,a process known as thermionic emission.
63
EasyMCQ
To emit an electron from a metal surface, the minimum electric field required is . . . . . . .
A
$10^{6} \,V \,m^{-1}$
B
$10^{8} \,V \,m^{-1}$
C
$10^{5} \,V \,m^{-1}$
D
$10^{4} \,V \,m^{-1}$

Solution

(B) The process of emitting electrons from a metal surface by applying a strong external electric field is known as field emission. For a typical metal, the work function is a few electron volts $(eV)$. To overcome this potential barrier and pull an electron out of the metal surface, an electric field of the order of $10^{8} \,V \,m^{-1}$ is required. Therefore, the correct option is $B$.
64
EasyMCQ
An electron is:
A
a hadron
B
a baryon
C
a nucleon
D
a lepton

Solution

(D) Leptons are a class of elementary particles that do not participate in strong nuclear interactions. They interact primarily through electromagnetic and weak nuclear forces.
An electron is a fundamental particle belonging to the lepton family.
65
EasyMCQ
The metal which has the highest work function among the following is:
A
Cesium $(Cs)$
B
Sodium $(Na)$
C
Aluminium $(Al)$
D
Platinum $(Pt)$

Solution

(D) Work function is the minimum energy required by an electron to escape from the metal surface. It depends on the properties of the metal and the nature of the surface.
Among the given options,Platinum $(Pt)$ has the highest work function,approximately $5.65 \ eV$,while Cesium $(Cs)$ has the lowest work function,approximately $2.14 \ eV$.
66
EasyMCQ
The minimum value of electric field required to pull out electrons from a metal is approximately . . . . . . $V$/cm.
A
$10^9$
B
$10^6$
C
$10^{10}$
D
$10^8$

Solution

(B) The electric field required to pull out an electron from a metal surface (a process known as field emission) is typically of the order of $10^6 \text{ V/cm}$.
This value is derived from the electrostatic force required to overcome the work function of the metal,allowing electrons to tunnel through the potential barrier.

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