Cathode Rays and Electron Emission Questions in English

(D) Cathode rays consist of a stream of negatively charged particles (electrons).
When a negatively charged particle enters an electric field,it experiences a force in the direction opposite to the electric field.
The electric field is directed from north to south.
Therefore,the force on the negatively charged cathode rays will be directed from south to north.
Thus,the deflection of the cathode rays is towards the north.

(D) The magnetic force acting on a charged particle moving in a magnetic field is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
Since the force $\vec{F}$ is always perpendicular to the velocity $\vec{v}$,the work done by the magnetic field on the particle is $W = \int \vec{F} \cdot d\vec{s} = \int \vec{F} \cdot \vec{v} dt = 0$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done,so the kinetic energy remains constant.
Since the speed of the particle remains constant,the magnitude of its momentum $(p = mv)$ also remains constant.
Therefore,both the energy and the magnitude of the momentum remain unchanged.

(B) strong argument for the particle nature of cathode rays is that they are deflected by electric and magnetic fields.
The deflection through electric and magnetic fields indicates that cathode rays are streams of minute particles carrying a negative charge.
Additionally,cathode rays cause mechanical motion of a small pinwheel placed in their path. Hence,they possess kinetic energy and must be material particles.

(B) The specific charge of a particle is defined as the ratio of its charge to its mass,given by $e/m$.
According to the theory of relativity,the mass $m$ of a particle moving with velocity $v$ is given by $m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$,where $m_0$ is the rest mass and $c$ is the speed of light.
As the speed $v$ of the electron increases,the relativistic mass $m$ increases.
Since the charge $e$ remains constant and the mass $m$ increases,the ratio $e/m$ (specific charge) decreases.

(C) Mass is generally considered a constant for physical applications at low velocities. However,according to Einstein's theory of special relativity,as an object's speed approaches the speed of light,its relativistic mass increases. The formula for the variation of mass with velocity is given by:
$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$
where $m$ is the relativistic mass,$m_0$ is the rest mass,$v$ is the velocity of the electron,and $c$ is the speed of light.

(B) Cathode rays consist of electrons,which are negatively charged particles.
When placed in a transverse electric field,these particles experience an electrostatic force $F = qE$.
Since the charge $q$ is negative,the force acts in the direction opposite to the electric field vector.
Electric field lines point from higher potential to lower potential.
Therefore,the force on the electrons acts from lower potential to higher potential,causing them to move up the potential gradient.

(B) In Thomson's experiment to determine the specific charge $(e/m)$ of an electron,a fine beam of electrons is passed through a region where both electric and magnetic fields are applied.
These fields are arranged such that they are mutually perpendicular to each other and also perpendicular to the direction of the electron beam.
By adjusting the fields,the electron beam is made to pass undeflected,allowing the determination of the velocity and the $e/m$ ratio.

(B) When a charged particle (like an electron in a cathode ray) enters a uniform magnetic field $B$ with a velocity $v$ perpendicular to the field,the magnetic Lorentz force $F = q(v \times B)$ acts as a centripetal force.
Since the force is always perpendicular to the velocity,the speed of the particle remains constant,but the direction changes continuously.
This results in the particle tracing a circular path.
Therefore,the correct option is $B$.

(C) The specific charge is defined as the ratio of charge $(e)$ to mass $(m)$ of a particle.
For an electron,the charge $e = 1.6 \times 10^{-19} \text{ C}$ and the mass $m = 9.1 \times 10^{-31} \text{ kg}$.
Therefore,the specific charge $\frac{e}{m} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 1.76 \times 10^{11} \text{ C/kg}$.

(D) Cathode rays consist of a stream of fast-moving electrons,while visible light consists of electromagnetic waves (photons).
Both cathode rays and visible light have the property of affecting photographic emulsions,meaning they can both expose a photographic plate.
Electric and magnetic fields deflect cathode rays (charged particles) but do not deflect light (neutral photons).
Only cathode rays can ionize a gas through collisions.
Therefore,the correct similarity is that they both can expose a photographic plate.

(C) $Cathode$ $ray$ $oscilloscope$ $(CRO)$ uses an electron beam generated by an electron gun. These high-speed electrons are directed towards a screen coated with a fluorescent material (like zinc sulfide). When the electrons strike the fluorescent screen,they transfer their kinetic energy to the atoms of the material,causing them to emit light,which is known as fluorescence. Therefore,the correct device is the $Cathode$ $ray$ $oscilloscope$.

(D) An oxide-coated filament is used in vacuum tubes because the oxide layer (typically barium or strontium oxide) significantly reduces the work function of the metal surface.
Due to this lower work function,the filament can emit a large number of electrons via thermionic emission at relatively lower temperatures compared to pure metal filaments.
This increases the efficiency and lifespan of the vacuum tube.

(A) In a discharge tube,cathode rays (a beam of negative particles) and canal rays (positive rays) move in opposite directions. The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B})$.
For positive rays,$q$ is positive and $\vec{v}$ is in one direction. For cathode rays,$q$ is negative and $\vec{v}$ is in the opposite direction.
When a magnetic field $\vec{B}$ is applied normally (perpendicular to the velocity),the cross product $\vec{v} \times \vec{B}$ changes sign for the cathode rays due to the negative charge,effectively resulting in a force in the same direction as that on the positive rays. Thus,they are deflected in the same direction.

(B) Cathode rays are streams of negatively charged particles known as electrons.
These rays are emitted from the cathode in a discharge tube under low pressure and high voltage.
They cause ionization of the gases through which they travel and are deflected by both electric and magnetic fields,confirming their negative charge.

(A) Cathode rays consist of a stream of fast-moving electrons.
These electrons possess significant kinetic energy due to their motion.
When these rays strike a metal plate,they collide with the atoms of the plate and transfer their kinetic energy to the plate.
This transfer of energy causes the atoms of the metal plate to vibrate more vigorously,resulting in an increase in the temperature of the plate.

(D) Cathode rays are streams of fast-moving electrons emitted from the cathode in a discharge tube.
Therefore,they are essentially beams of electrons.
Thus,the correct option is $D$.

(C) Cathode rays are a stream of fast-moving electrons. They are negatively charged particles and can be deflected by both electric and magnetic fields. However,their speed ranges from $10^7 \, m/s$ to $3 \times 10^7 \, m/s$,which is significantly less than the speed of light $(c \approx 3 \times 10^8 \, m/s)$. Therefore,statement $(c)$ is not true.

(C) Cathode rays are produced in a discharge tube at low pressure. The typical pressure required for the production of cathode rays is approximately $0.01 \ mm$ of $Hg$ (or $10^{-2} \ mm$ of $Hg$).
At this low pressure,the mean free path of the electrons becomes large enough to allow them to accelerate towards the anode without frequent collisions with gas molecules.
If the pressure is higher,the number of gas atoms increases,leading to more frequent collisions between electrons and gas atoms,which obstructs the flow of electrons and prevents the formation of a stable cathode ray beam.

(D) Cathode rays consist of a stream of negatively charged particles (electrons). Because they carry a charge,they are deflected by both electric and magnetic fields. Therefore,the statement that they do not deflect in an electric field is incorrect.

(C) When electrons pass through crossed electric $(E)$ and magnetic $(B)$ fields undeflected,the electric force equals the magnetic force: $eE = evB$,which implies $v = E/B$.
The kinetic energy gained by an electron accelerated through a potential difference $V$ is given by $\frac{1}{2}mv^2 = eV$,so $v^2 = \frac{2eV}{m}$.
Substituting $v = E/B$ into the equation,we get $(E/B)^2 = \frac{2eV}{m}$,which simplifies to $\frac{e}{m} = \frac{E^2}{2VB^2}$.
Given values: $E = 3.6 \times 10^4 \ Vm^{-1}$,$B = 1.2 \times 10^{-3} \ T$,$V = 2.5 \times 10^3 \ V$.
Calculating $\frac{e}{m} = \frac{(3.6 \times 10^4)^2}{2 \times (2.5 \times 10^3) \times (1.2 \times 10^{-3})^2} = \frac{12.96 \times 10^8}{5000 \times 1.44 \times 10^{-6}} = \frac{12.96 \times 10^8}{7.2 \times 10^{-3}} = 1.8 \times 10^{11} \ C \ kg^{-1}$.

(B) In a gas discharge tube,the appearance of different regions depends on the pressure of the gas.
At a pressure of approximately $0.01\, mm$ to $0.05\, mm$ of $Hg$,the discharge tube exhibits the Crooke's dark space $(CDS)$,which occupies almost the entire tube.
Since $0.02\, mm$ falls within this range,the formation of Crooke's dark space occurs.

(B) The current $i$ is defined as the rate of flow of charge,given by $i = \frac{q}{t} = \frac{ne}{t}$.
Given,the number of electrons emitted per second $\frac{n}{t} = 1.8 \times 10^{14} \ s^{-1}$.
The charge of an electron $e = 1.6 \times 10^{-19} \ C$.
Substituting the values:
$i = (1.8 \times 10^{14}) \times (1.6 \times 10^{-19}) \ A$
$i = 2.88 \times 10^{-5} \ A$
$i = 28.8 \times 10^{-6} \ A$
Since $1 \ \mu A = 10^{-6} \ A$,we have $i = 28.8 \ \mu A$.
Rounding to the nearest integer,the maximum anode current is approximately $29 \ \mu A$.

(A) The specific charge is defined as the ratio of charge to mass $(q/m)$.
For an electron: $q_e = e$,$m_e \approx 9.1 \times 10^{-31} \ kg$.
For a proton: $q_p = e$,$m_p \approx 1.67 \times 10^{-27} \ kg$.
For an $\alpha$-particle: $q_{\alpha} = 2e$,$m_{\alpha} \approx 4 \times 1.67 \times 10^{-27} \ kg$.
Calculating the ratios:
$(q/m)_e = e / m_e$
$(q/m)_p = e / m_p$
$(q/m)_{\alpha} = 2e / (4m_p) = 0.5 (e / m_p)$
Since $m_e \ll m_p$,the ratio for the electron is the largest.
Comparing proton and $\alpha$-particle: $(q/m)_p > (q/m)_{\alpha}$.
Therefore,the order is $(q/m)_e > (q/m)_p > (q/m)_{\alpha}$.

(C) Positive rays,also known as canal rays,are streams of positively charged particles produced in a discharge tube. These particles are formed when high-energy electrons collide with gas atoms,stripping away electrons and leaving behind positive ions. Therefore,positive rays consist of positive ions.

(B) Positive rays,also known as canal rays,were discovered by $E. Goldstein$ in $1886$ using a discharge tube with a perforated cathode.
However,the detailed study and identification of these rays as positively charged particles (ions) were later performed by $J.J. Thomson$.
In the context of standard physics curriculum questions,$E. Goldstein$ is credited with the discovery of canal rays (positive rays).

(A) Specific charge is defined as the ratio of charge to mass $(q/m)$.
For a positron, the charge is $+e$ and the mass is $m_e$ (mass of an electron).
Specific charge of positron = $e/m_e$.
For a proton, the charge is $+e$ and the mass is $m_p \approx 1836 m_e$.
Specific charge of proton = $e/m_p = e/(1836 m_e)$.
For an $He$ nucleus (alpha particle), the charge is $+2e$ and the mass is $m_{\alpha} \approx 4 m_p \approx 7344 m_e$.
Specific charge of $He$ nucleus = $2e/(7344 m_e) = e/(3672 m_e)$.
Comparing the values, the positron has the smallest mass, therefore it has the highest specific charge.

(C) photoelectron is essentially an electron that has been emitted from a material due to the photoelectric effect.
Since it is an electron,its mass is equal to the rest mass of an electron.
The rest mass of an electron is approximately $9.1 \times 10^{-31} \ kg$.
Therefore,the correct option is $C$.

(C) According to the Richardson-Dushman equation,the thermionic emission current density $J$ is given by:
$J = AT^2 e^{-W_0/kT}$
where $A$ is the Richardson constant,$T$ is the absolute temperature,$W_0$ is the work function of the metal,and $k$ is the Boltzmann constant.
From this equation,it is evident that the emission current density $J$ is proportional to the square of the absolute temperature,i.e.,$J \propto T^2$.
Therefore,the correct option is $C$.

(A) The ratio of the number of secondary electrons emitted to the number of primary electrons incident is known as the secondary emission coefficient $\delta$.
This coefficient $\delta$ is primarily a property of the target material and the energy of the primary electrons.
However, in the context of standard physics problems regarding electron emission, if the question asks what determines the total number of secondary electrons emitted, it is directly proportional to the intensity (number of incident primary electrons) of the primary beam.
Given the options provided, the secondary emission coefficient itself is a material property, but the total count of secondary electrons depends on the intensity of the primary beam.
Therefore, the most appropriate answer is (A) Material of the target, as $\delta$ is a characteristic constant for a given material.

(C) The given equation is $I = A T^2 e^{qV / V_L}$.
Comparing this with the standard Richardson-Dushman equation for thermionic emission,$I = A T^2 e^{-qV / kT}$ (where the exponent term represents the ratio of energy to thermal energy).
In the given expression,the exponent is $\frac{qV}{V_L}$.
Comparing the exponents: $\frac{qV}{V_L} = \frac{qV}{kT}$.
Therefore,$V_L = \frac{kT}{V}$.

(B) The thermionic emission is described by the Richardson-Dushman equation.
According to this equation,the current density $J$ emitted from a heated surface is given by $J = AT^2 e^{-\varphi/kT}$,where $A$ is the Richardson constant,$T$ is the absolute temperature,$\varphi$ is the work function of the material,and $k$ is the Boltzmann constant.
Therefore,the correct relation is $J = AT^2 e^{-\varphi/kT}$.

(C) Thermionic emission is the process by which electrons are emitted from the surface of a metal due to the absorption of thermal energy. When a metal is heated to a high temperature,the free electrons within the metal gain sufficient kinetic energy to overcome the work function of the metal surface and escape into the surrounding space. Therefore,thermionic emission is caused by high temperature.

(C) The emission current $i$ is given by the Richardson-Dushman equation: $i = A T^2 S e^{-\phi / kT}$.
Given for the first surface: $\phi_1 = 2 \, eV$,$T_1 = 800 \, K$,$i_1 = 1 \, mA$.
Given for the second surface: $\phi_2 = 4 \, eV$,$T_2 = 1600 \, K$,$A_1 = A_2$,$S_1 = S_2$.
We observe that the ratio $\frac{\phi_1}{T_1} = \frac{2}{800} = 0.0025$ and $\frac{\phi_2}{T_2} = \frac{4}{1600} = 0.0025$. Thus,$\frac{\phi_1}{T_1} = \frac{\phi_2}{T_2}$.
Taking the ratio of the currents: $\frac{i_2}{i_1} = \frac{A T_2^2 S e^{-\phi_2 / kT_2}}{A T_1^2 S e^{-\phi_1 / kT_1}} = \left( \frac{T_2}{T_1} \right)^2 \cdot e^{-(\phi_2/kT_2 - \phi_1/kT_1)}$.
Since $\frac{\phi_2}{kT_2} = \frac{\phi_1}{kT_1}$,the exponential term becomes $e^0 = 1$.
Therefore,$\frac{i_2}{i_1} = \left( \frac{1600}{800} \right)^2 = (2)^2 = 4$.
Hence,$i_2 = 4 \times i_1 = 4 \times 1 \, mA = 4 \, mA$.

(A) According to the Richardson-Dushman equation,the current density $J$ is given by $J = A T^2 e^{-b/T}$,where $A$ and $b$ are constants.
Taking the natural logarithm on both sides:
$\log_e J = \log_e A + 2 \log_e T - \frac{b}{T}$
Rearranging the terms to isolate $\log_e \frac{J}{T^2}$:
$\log_e \frac{J}{T^2} = \log_e A - \frac{b}{T}$
This equation is of the form $y = mx + c$,where $y = \log_e \frac{J}{T^2}$,$x = \frac{1}{T}$,$m = -b$ (slope),and $c = \log_e A$ (intercept).
Since the slope $m = -b$ is negative,the graph is a straight line with a negative slope and a positive intercept on the $y$-axis.
Therefore,the correct curve is represented by the graph with a downward-sloping straight line.

(C) The Richardson-Dushman equation for thermionic emission is given by $J = AT^2 e^{-b/T}$,where $A$ and $b$ are constants.
Dividing both sides by $T^2$,we get $\frac{J}{T^2} = A e^{-b/T}$.
Taking the natural logarithm on both sides,we get $\ln\left(\frac{J}{T^2}\right) = \ln(A) - \frac{b}{T}$.
This equation is of the form $y = mx + c$,where $y = \ln\left(\frac{J}{T^2}\right)$,$x = \frac{1}{T}$,$m = -b$ (slope),and $c = \ln(A)$ (intercept).
However,if we plot $\frac{J}{T^2}$ directly against $\frac{1}{T}$,the relationship is $\frac{J}{T^2} = A e^{-b/T}$.
As $\frac{1}{T}$ increases,the exponent $-b/T$ becomes more negative,causing the value of $\frac{J}{T^2}$ to decrease exponentially towards zero.
Therefore,the curve is an exponential decay curve.

(A) According to Richardson-Dushman equation,the thermionic current $I$ is given by $I = AT^2 e^{-\phi/kT}$,where $A$ is a constant,$T$ is the absolute temperature,$\phi$ is the work function,and $k$ is the Boltzmann constant.
For a small range of temperature variation,the exponential term $e^{-\phi/kT}$ remains approximately constant.
Therefore,the current $I$ is proportional to the square of the temperature: $I \propto T^2$.
This implies the ratio $\frac{I}{I_0} = \left( \frac{T}{T_0} \right)^2$.
This equation represents a parabola opening upwards,which corresponds to curve $A$ in the given figure.

(C) . Cathode rays are not electromagnetic in nature.
Cathode rays (electron beam or $e$-beam) are streams of electrons observed in vacuum tubes. If an evacuated glass tube is equipped with two electrodes and a voltage is applied,the glass behind the positive electrode is observed to glow due to electrons emitted from and traveling away from the cathode (the electrode connected to the negative terminal of the voltage supply). They were first observed in $1869$ by German physicist Johann Wilhelm Hittorf and were named in $1876$ by Eugen Goldstein as $Kathodenstrahlen$,or cathode rays. In $1897$,British physicist $J.J. Thomson$ showed that cathode rays were composed of a previously unknown negatively charged particle,which was later named the electron. Cathode ray tubes $(CRTs)$ use a focused beam of electrons deflected by electric or magnetic fields to create the image on a television screen.
The electromagnetic force usually exhibits electromagnetic fields such as electric fields,magnetic fields,and light,and is one of the four fundamental interactions in nature. The other three fundamental interactions are the strong interaction,the weak interaction,and gravitation.

(C) Thermions are electrically charged particles or ions that are emitted by a heated conducting material.
When a metal surface is heated to a high temperature,the thermal energy provided to the free electrons allows them to overcome the work function of the metal and escape the surface.
These emitted electrons are specifically referred to as thermions.
Therefore,thermions are electrons.

(B) The elementary charge of an electron was accurately measured by Robert $A$. Millikan in his famous oil-drop experiment conducted in $1909$. While $J$. $J$. Thomson discovered the electron itself,Millikan is credited with determining the magnitude of the charge on an electron.

(D) Cathode rays are streams of fast-moving electrons emitted from the cathode in a discharge tube.

(B) In a gas discharge tube,the color of the positive column is determined by the nature of the gas present inside the tube.
Different gases emit light of different characteristic wavelengths when excited by an electric discharge.
For example,air produces a pinkish-red glow,while hydrogen $(H_2)$ produces a blue glow.

(B) In an electron gun,the cathode emits electrons via thermionic emission. The control grid is placed between the cathode and the accelerating anode. By maintaining the control grid at a negative potential relative to the cathode,it acts as a gate that repels the electrons. This allows for the regulation of the number of electrons passing through the aperture,thereby controlling the intensity of the electron beam.

(D) When a beam of cathode rays (electrons) is subjected to crossed electric $(E)$ and magnetic $(B)$ fields,the beam is not deflected if the force due to the electric field equals the force due to the magnetic field.
$Be v = eE$
$v = \frac{E}{B}$ ..... $(i)$
If $V$ is the potential difference between the anode and the cathode,the kinetic energy gained by the electron is given by:
$\frac{1}{2} m v^2 = eV$
$\frac{e}{m} = \frac{v^2}{2V}$ ..... $(ii)$
Substituting the value of $v$ from equation $(i)$ into equation $(ii)$,we get:
$\frac{e}{m} = \frac{(E/B)^2}{2V} = \frac{E^2}{2VB^2}$
Thus,the specific charge of the cathode rays is $\frac{E^2}{2VB^2}$.

(C) In Thomson's experiment,the deflection of a charged particle in an electric or magnetic field depends on the specific charge,which is the ratio of charge to mass,denoted as $\frac{q}{m}$.
Since all three particles produce the same spot on the screen,they must experience the same deflection,which implies that their specific charge ratios must be equal.
Therefore,$\frac{q_1}{m_1} = \frac{q_2}{m_2} = \frac{q_3}{m_3}$.
Given the ratio of charges is $q_1 : q_2 : q_3 = 1 : 3 : 5$.
For the ratios to be equal,the masses must follow the same ratio as the charges.
Thus,$m_1 : m_2 : m_3 = 1 : 3 : 5$.

(B) $X$-rays emission: These are produced due to transitions in the inner energy levels of an atom,not by the emission of electrons from the surface.
Photoelectric emission: This is the emission of electrons from a metal surface upon irradiation with radiation of a suitable frequency.
Secondary emission: When an incident electron strikes the surface of a metallic plate,it causes the emission of other electrons from the surface.
Thermionic emission: When a metal is heated to a high temperature,the free electrons gain sufficient kinetic energy to escape from the surface of the metal.

(B) $J$.$J$. Thomson determined the specific charge $(e/m)$ of an electron using his cathode ray tube experiment. Later,Robert $A$. Millikan determined the charge $(e)$ of an electron using his oil-drop experiment. By combining these two values,the mass $(m)$ of the electron was evaluated indirectly as $m = e / (e/m)$.

(B) In the phenomenon of electric discharge through gases at low pressure,a high potential difference is applied across the electrodes.
This causes the emission of charged particles (electrons) from the cathode.
These high-speed electrons collide with the neutral gas atoms present in the tube.
These collisions cause the excitation and ionization of the gas atoms.
When these excited atoms return to their ground state,they emit light of specific wavelengths,which results in the characteristic coloured glow observed in the discharge tube.

(A) photoemissive cell consists of two electrodes enclosed in a glass bulb,which may be evacuated or filled with an inert gas at low pressure.
When an inert gas is present,the photoelectrons emitted from the cathode collide with the gas atoms,causing ionization. This process increases the number of charge carriers,resulting in a greater current compared to an evacuated cell.
Therefore,both the Assertion and the Reason are correct,and the Reason correctly explains why an inert gas is used in such cells.

(A) Potential difference $V = 500\;V$. Specific charge $e/m = 1.76 \times 10^{11}\;C\;kg^{-1}$.
Using the conservation of energy,$eV = \frac{1}{2}mv^2$,we get $v = \sqrt{2V(e/m)}$.
$v = \sqrt{2 \times 500 \times 1.76 \times 10^{11}} = \sqrt{1.76 \times 10^{14}} \approx 1.33 \times 10^7\;m/s$.
$(b)$ For $V = 10\;MV = 10^7\;V$,the formula gives $v = \sqrt{2 \times 10^7 \times 1.76 \times 10^{11}} = \sqrt{3.52 \times 10^{18}} \approx 1.88 \times 10^9\;m/s$.
This result is wrong because $v > c$ (speed of light),which is physically impossible.
The formula $K.E. = \frac{1}{2}mv^2$ is non-relativistic. For high potentials,we must use the relativistic kinetic energy formula: $K.E. = (\gamma - 1)mc^2$,where $\gamma = (1 - v^2/c^2)^{-1/2}$.

(A) In $1870$,William Crookes performed experiments with a gas discharge tube.
He filled a gas in the discharge tube at low pressure and applied a high electric field to the electrodes at both ends of the tube.
He observed that electric rays (cathode rays) were produced,which traveled from the cathode to the anode. When these rays struck a fluorescent material,they produced a yellowish glow.
$J$.$J$. Thomson further investigated these rays and observed that they could be deflected by electric and magnetic fields,confirming they carry a negative charge.
$J$.$J$. Thomson measured the speed of cathode rays,finding it to be in the range of $\frac{1}{20}$ to $\frac{1}{30}$ times the speed of light $(3 \times 10^{8} \ m/s)$.
He determined the specific charge (ratio of charge to mass,$\frac{e}{m}$) of these particles to be approximately $1.76 \times 10^{11} \ C/kg$.
He concluded that the specific charge $\frac{e}{m}$ is independent of the type of gas used in the tube and the material of the electrodes,suggesting that these particles (electrons) are universal constituents of all matter.