## Solved Problems

Here are a few examples to help you understand the how to calculate the work function of a metal.

## What is Work Function?

**Definition of work function:** The work function of a metal is the minimum amount of energy that is needed to eject an electron from the surface of a metal. The unit of work function electronvolt (eV) or Joules(J). The symbol for the work function is Φ or W.

The work function of a metal measures the minimum amount of energy that can liberate an electron from the surface of that metal.

## Work Function Formula

The formula for calculating the work function are:

- The first formula for calculating work function is Φ = E + K.E [where Φ = work function, E = total energy, and K.E = Maximum kinetic energy of liberated photoelectrons]
- We also have Φ = hf
_{0}[where w = work function of the metal, h=Planck’s constant, and f_{0}=Threshold frequency] - The third formula for work function is Φ = hc/λ [where h=planck’s constant, c= speed of light in the vacuum, and λ=wavelength of the radiation]. This is because f=c/λ, and since Φ = hf, this implies that Φ = hc/λ

## Photoelectric Equation

Photoelectric effect is due to the emission of electrons from the surface of a metal when light is shown on it. We can write photoelectric equation as

E = hf = Φ + (1/2)mv^{2}

Where

E = energy required to liberate an electron from the surface of a metal

h = Planck’s constant

f = this is the frequency

Φ = this symbol represents the work function

m = mass of an electron

v = speed of the electron

### Example 1

A metal is illuminated with radiation of energy 6.88eV. If the kinetic energy of the emitted electrons is 1.5eV. Calculate the work function of the metal.

**Text** **Solution**

**Data:**

The work function, Φ = Total energy + Kinetic energy

and the the incident radiation = total energy = 6.8eV

The kinetic energy of the emitted electron = 1.5eV

and the formula is Φ = E + K.E

Thus,

Work function, Φ = 6.8eV + 1.5eV = 8.4eV

### Example 2

A metal was exposed to photons at a frequency of 2.0 x 10^{15}Hz, and electrons were emitted with a maximum kinetic energy of 4.00 x 10^{-19} J. Calculate the work function, of this metal.

#### Text **Solution:**

**Data:**

Φ = hf + K.E

and

Frequency of photons, f = 2.0 x 10^{15}Hz

The maximum kinetic energy of the electron, K.E = 4.00 x 10^{-19} J

planck’s constant, h = 6.6×10^{-34}Js

Hence, we can now apply the formula for the work function that says Φ = hf + K.E

Thus, we now have Φ = (6.6×10^{-34}) x (2.0 x 10^{15}) + (4.00 x 10^{-19})

This is equal to

Φ = (6.6×10^{-34}) x (2.0 x 10^{15}) + (4.00 x 10^{-19})

we now have

Φ = 1.32×10^{-18} + 4.00 x 10^{-19}

Therefore, the work function Φ, of the metal is 1.72 x 10^{-18} J

### Example 3

When metal was exposed to photons at a frequency of 1.10 x 10^{15}Hz, electrons were emitted with a maximum kinetic energy of 4.00 x 10^{-19}J. Calculate the work function of the metal.

**Text Solution**

**Data:**

frequency, f = 1.10 x 10^{15}Hz

Maximum kinetic energy, K.E = 4.00 x 10^{-19}J

The work function of the metal, Φ =?

Planck’s constant, h = 6.6×10^{-34}Js

Therefore, by applying the formula for the work function that says Φ = hf + K.E

After substituting our data into the above formula for the work function, we will now have

Φ = (6.6×10^{-34}) x (1.10 x 10^{15}) + (4.00 x 10^{-19})

After multiplying (6.6×10^{-34}) by (1.10 x 10^{15}) the above expression will give us

Φ = (7.26×10^{-19}) + (4.00 x 10^{-19})

We will now add (4.00 x 10^{-19}) to (7.26×10^{-19}), Which implies that the above expression is now

The work function of the metal, Φ = 1.126×10^{-19}J

**Therefore, the work function of the metal Φ is 1.126×10 ^{-19}J**

### Example 4

Calculate the work function of a metal in eV, if its threshold wavelength is 6,800Å [Take h = 6.6 x 10^{-27} ergs].

#### Text **Solution**

**Data:**

The wavelength, λ = 6,800Å

and from our knowledge of prefixes, the symbol Å is called angstrom, and 1Å = 0.0000000001 or 10^{-10}m

Thus, the wavelength, λ = 6,800Å = 6,800 x 10^{-10}m = 6.86 x 10^{-6}m

Therefore, λ = 6.86 x 10^{-6}m

From the above question, we were told that Planck’s constant, h = 6.6 x 10^{-27} ergs = 6.6 x 10^{-27} x 10^{-7 }[This is because 1 erg is equal to 10^{-7}J]

Therefore we can now write Planck’s constant, h = 6.6 x 10^{-27} x 10^{-7} = 6.6 x 10^{-27-7} = 6.6 x 10^{-34}Js

##### Solution (applying data and the work function formula)

We will apply the formula for the work function, which is Φ = hc/λ [where h=planck’s constant, c= speed of light in the vacuum, and λ=wavelength of the radiation]

Speed of light in the vacuum, c = 3 x 10^{8} m/s

After substituting our values into the above formula, we will have

Φ = (6.6 x 10^{-34} x 3 x 10^{8})/6.86 x 10^{-6}

We multiply the numerator (6.6 x 10^{-34} x 3 x 10^{8}) to obtain 1.98 x 10^{-25}

Thus,

Work function, Φ = 1.98 x 10^{-25}/6.86 x 10^{-6}

Dividing 1.98 x 10^{-25} by 6.86 x 10^{-6} will give us 2.886 x 10^{-20}

Work function, Φ = 2.886 x 10^{-20}J

**Which explains that the work function of the metal is 2.886 x 10 ^{-20}J**

To convert our answer into electronvolt, we consider 1eV = 1.6 x 10^{-19}J

Hence, Work function, Φ = (2.886 x 10^{-20}J x 1eV) / 1.6 x 10^{-19}J = 0.179125eV

Therefore, our work function in electronvolt, Φ = 0.179125eV or approximately into 0.2eV

## Explanation

Understanding the work function is crucial in various fields, including semiconductor devices, solar cells, and vacuum electronics. Thus, the energy from work function can be supplied through thermal excitation, electrical bias, or electromagnetic radiation. In essence, the work function is the energy required to overcome the potential barrier that holds the electrons within the material’s surface. Additionally, the work function is measured in electron volts (eV) and is denoted by the symbol Φ.

### The Importance of the Work Function

Work function is a crucial parameter in determining the electronic properties of materials. For instance, we use work function to understand the electronic behavior of metals, semiconductors, and insulators. Moreover, work function is also a crucial parameter in determining the efficiency of electronic devices. Some of these devices are photovoltaic cells, field-effect transistors, and thermionic converters.

### Factors Affecting the Work Function

It is also important to know that several factors can influence the work function of a material. These factors include:

- The surface orientation,
- Crystallographic structure,
- Temperature, and
- Doping.

For instance, the work function of a metal surface may vary depending on the **crystallographic orientation** of the surface. **Doping**, on the other hand, involves the intentional introduction of impurities into the material. Hence, this action can alter the electronic properties of the material, including the work function.

### Calculating the Work Function

We can apply different approach to calculate the work function of a material. Some of these approaches include:

#### The Photoelectric Effect

The photoelectric effect involves the emission of electrons from a material’s surface when exposed to electromagnetic radiation. Therefore, we can use the energy of the incident radiation to determine the work function of the material.

E = hf = Φ + (1/2)mv^{2}

#### The Richardson-Dushman Equation

When we talk of Richardson-Dushman equation. We are relating the current density of a heated metal surface to the temperature and the work function. This equation can be used to determine the work function of a metal. The mathematical expression of Richardson-Dushman equation is

J_{e} = A T^{2} exp(-Φ/kT)

Where

A = Richardson -Dushman Constant

J_{e} = Current density due to the emission

k = Boltzman constant

T = temperature of the emission

Φ = work function

#### The Kelvin Probe Method

The Kelvin probe method involves measuring the potential difference between a metal surface and a reference electrode. This potential difference can be used to determine the work function of the metal.

C=Q/V=(epsilon. A)/t

C = capacitance

Q = charge

V = Potential difference

epsilon = Permittivity of the dielectric

A = Surface area of the capacitor

t = time taken

### Applications of the Work Function

The work function formula finds numerous applications in various fields. For instance, we use the work function to understand the electronic properties of materials in the semiconductor industry. Additionally, It is also a crucial parameter in determining the efficiency of photovoltaic cells. In vacuum electronics, the work function is used to understand electron emission from hot cathodes.

### Frequently Asked Questions (FAQs)

**What is the work function?**

The work function formula refers to the energy needed to remove an electron from a solid material’s surface and is measured in electron volts (eV).

**What factors affect the work function of a material?**

Several factors can influence the work function of a material, including surface orientation, temperature, doping, and crystallographic structure.

**What is the work function and threshold frequency formula?**

The formula for work function and threshold frequency is **E = hf = Φ + (1/2)mv ^{2}**

**What is the work function unit?**

The s.i unit of the work function is in Joules

**What is work function formula in Terms of wavelength**?

The work function formula in Terms of wavelength is **Φ = hc/λ**

we can also use the formula **E = hc/λ + (1/2)mv ^{2} **

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**Reference**