Electron Emission

Electron emission refers to the process by which electrons are released from the surface of a material into its surrounding environment. This phenomenon occurs in several ways, including photoelectric emission, thermionic emission, and field emission. In photoelectric emission, electrons are ejected when light hits the material, a principle utilized in devices like solar panels and photoelectric sensors. Thermionic emission involves the release of electrons from a heated material, which is crucial in the operation of vacuum tubes and cathode ray tubes. Field emission occurs when a strong electric field pulls electrons from a material, an effect used in technologies like electron microscopes and certain types of display screens. In this article, we will discuss the concept of electron emission and provide examples for better understanding.

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What is Electron Emission?

As we have learned in Chemistry (Atomic structure) the electrons in the outermost orbit of an atom are at maximum distance from the nucleus and hence most loosely bound to it. This type of electron is called free electrons. The free electrons in metals are free to move within the volume of metal even though they do not get ejected out of the surface of metal on their own. The main reason behind this is that whenever an electron tries to leave the surface, the surface acquires a positive charge which pulls back the electron. So for escaping from the surface, an electron has to do a definite amount of work to overcome the force exerted by the opposite charges. To do this work, an external source imparts minimum energy. This minimum energy is called the work function of the metal and is denoted by \phi_0.

The work function of any particular material is defined as the minimum energy which is required to liberate the most weakly bound surface electrons from that material without giving them any velocity. Since it is energy, it is generally denoted in electron volt (eV).

$\mathrm{leV}=\mathrm{le} \times 1 \mathrm{~V}=\left(1.6 \times 10^{-19} \mathrm{C}\right)(1 \mathrm{~V})=1.6 \times 10^{-19} \mathrm{~J}$

Since the energy of a photon is given by $h \nu$, So the minimum energy i.e., the Work function is given by

$
\phi=h \nu_o=\frac{h c}{\lambda_o}
$

When a free electron gets extra energy i.e., imparted energy $\geq$ work function from an external agent, then it is able to overcome the potential barrier and the electron gets ejected out. There are a number of ways in which energy from outside can be supplied and based on this different way, there are different ways in which electron emission can take place. These ways are listed below:

1. Photoelectric emission: When electromagnetic radiations of suitable frequency (or wavelength) are incident on a metallic surface, then electrons will be emitted, this phenomenon is known as the photoelectric effect.

2. Thermionic emission: In this case, additional energy is given to the electrons to overcome a potential barrier in the form of heat by passing current through a filament.

3. Field emission: In this case, metal is placed in a strong electric field due to which the electrons are accelerated to such a speed that the corresponding kinetic energy is sufficient to overcome the potential barrier.

4. Secondary emission: It is a process in which the work function is supplied to the free electrons of a metal surface by collisions with fast-moving secondary particles like neutrons, beta particles, etc.

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Solved Examples Based on Dual Nature of Matter and Radiation

Example 1: 5v of stopping potential needed for the photoelectrons emitted out of a surface of work function 2.2ev by the radiation of wavelength :

1) $1719 \mathrm{~A}^{\circ}$
2) $8.3444 A^{\circ}$
3) $861 A^{\circ}$
4) $3000 A^{\circ}$

Solution:

Stopping Potential /Cut-off Potential

It is defined as the potential necessary to stop any electron from reaching the other side.

$\begin{aligned} & h \nu=5 e v+2.2 e v=7.2 e v \\ & 7.2=\frac{h c}{\lambda A^{\circ}} \Rightarrow \lambda=\frac{h c}{7.2(e v)}=\frac{12400}{7.2} A^{\circ} \\ & \lambda=1719 A^{\circ}\end{aligned}$

Hence, the correct answer is option (1).

Example 2: When a certain photosensitive surface is illuminated with monochromatic light of frequency $\nu$, the stopping potential for the photo-current is $-\frac{V_0}{2}$. When the surface is illuminated by monochromatic light of frequency $\overline{2}$, the stopping potential is $-V_0$. The threshold frequency for photoelectric emission is :

1) $2 \nu$
2) $\frac{3 \nu}{2}$
3) $\frac{4 \nu}{3}$
4) $\frac{5 \nu}{3}$

Solution:

Stopping Potential /Cut-off Potential $V_0$
It is defined as the potential necessary to stop any electron from reaching the other side.
Stopping Potential is related to maximum kinetic energy

$
K_{\max }=\left|e V_0\right|
$

and Einstein's Photoelectric equation is given as

$
\mathrm{K}_{\max }=\mathrm{hv}-w
$

For $v$ frequency stopping potential is $\frac{V_0}{2}$
For $\frac{\nu}{2}$ frequency stopping potential is $-V_0$
So for $\nu$ frequencyl

$
h \nu=w+\frac{V_0}{2} e_{\ldots \ldots .1}
$

$
h \nu=w+\frac{V_0}{2} e_{\ldots \ldots .1}
$

For $\frac{\nu}{2}$ frequency

$
h \frac{\nu}{2}=w+V_0 e \ldots \ldots .2
$

From 1 and 2
We get

$
\begin{aligned}
& w=\frac{3}{2} h \nu \\
& w=h v_0=\frac{3}{2} h v
\end{aligned}
$

so

$
v_0=\frac{3}{2} v
$

Hence, the correct option is 2.

Example 3: In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm . The decrease in the stopping potential (in V) is close to:

$
\left(\frac{h c}{e}=1240 n m-V\right)
$

1) 1.0

2) 0.5

3) 1.5

4) 2.0

Solution:

For 1

$
\frac{h c}{\lambda_1}=\phi+e V_1
$
For 2

$
\frac{h c}{\lambda_2}=\phi+e V_2 \ldots \cdots
$
$
\begin{aligned}
& \text { (2) - (1) } \\
& \Rightarrow h C\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right)=e\left(V_2-V_1\right) \\
& \left(\frac{h c}{e}\right)\left(\frac{\lambda_2-\lambda_1}{\lambda_2 \lambda_1}\right)=V_2-V_1 \\
& \text { So, } V_2-V_1=\frac{(400-300)}{400 \times 300} \times 1240 \\
& =\frac{100}{400 \times 300} \times 1240 \\
& V_2-V_1 \approx 1 V
\end{aligned}
$

Hence, the answer is 1.0.

Example 4: Two identical photocathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photoelectron (of mass m ) coming out are $v_1$ and $v_2$ respectively, then the correct relation among the following is:

${ }_{1)} v_1-v_2=\left[\frac{2 h}{m}\left(f_1-f_2\right)\right]^{1 / 2}$
2) $V_1^2-V_2^2=\frac{2 h}{m}\left(f_1-f_2\right)$
3) $v_1^2+v_2^2=\frac{2 h}{m}\left(f_1-f_2\right)$
4) $v_1+v_2=\left[\frac{2 h}{m}\left(f_1-f_2\right)\right]^{1 / 2}$

Solution:

Einstein’s Photoelectric equation can also be written as

$\begin{aligned} & \mathrm{K} . \mathrm{E} .=\mathrm{hf}-\Phi \\ & h f_1-h f_0=\frac{1}{2} m v_1^2 \\ & h f_2-h f_0=\frac{1}{2} m v_2^2 \\ & \Rightarrow h\left(f_1-f_2\right)=\frac{1}{2} m\left(v_1^2-v_2^2\right) \\ & \Rightarrow v_1^2-v_2^2=\frac{2 h}{m}\left(f_1-f_2\right)\end{aligned}$

Hence, the correct answer is option (2).

Example 5: The time taken by a photoelectron to come out after the photon strikes is approximately

1) 10^-1 s

2) 10^-4 s

3) 10^-10 s

4) 10^-16 s

Solution:

Time taken is of the order of nanosecond i.e. around 10^-10 sec.

Hence, the answer is the option (3).

Summary

Electron emission encompasses various processes where electrons are expelled from a material's surface. Key methods include photoelectric emission, where light induces electron release, thermionic emission, which involves heating to release electrons, and field emission, driven by strong electric fields. These principles are pivotal in technologies such as solar panels, vacuum tubes, and electron microscopes. Examples illustrate calculations involving stopping potentials, frequency effects, and energy requirements, highlighting the practical implications and applications of these phenomena in scientific and technological contexts.