## What is 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.

## Video Explanation: How to Calculate the Work Function

Watch the video below for more explanation and a few solved problems:

## Work Function Formula and Video Explanation

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

To comprehend the concept of work function better, We need to look into photoelectric effect. The photoelectric effect occurs when light, typically in the form of photons, strikes the surface of a material and causes the ejection of electrons.

**Definition of Photoelectric Effect:** The photoelectric effect is due to the emission of electrons from the surface of a metal when light is shown on it. We can write the 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

## How to Calculate the Work Function of a Metal

**Calculating the Work Function**

To calculate the work function, you need to follow a step-by-step process:

**Step 1: Determine the Energy of the Incident Photon**

The energy of a photon is given by the equation E = hf, where E represents energy, h is Planck’s constant (approximately 6.626 x 10^{-34} J·s), and f denotes the frequency of the photon.

**Step 2: Calculate the Kinetic Energy of the Ejected Electron**

The kinetic energy (KE) of the ejected electron can be determined using the equation KE = hf – φ, where φ is the work function of the material.

**Step 3: Calculate the Work Function**

Rearranging the equation from Step 2, we can determine the work function as φ = hf – KE.

## Applicable Methods of Finding Work Function of a Metal

**Method 1: Calculating the Work Function of a Metal Surface**

Suppose we have a metal surface that is struck by a photon with a frequency of 2.5 x 10^{14} Hz. The kinetic energy of the ejected electron is measured as 4.0 x 10^{-19} J. By substituting the given values into the equation φ = hf – KE, we can calculate the work function.

**φ = hf – KE = (6.626 x 10 ^{-34} x 2.5 x 10^{14}) – 4.0 x 10^{-19} = -2.3435 x 10^{-19}**

**Example 2: Calculating the Work Function of a Semiconductor**

In the case of a semiconductor material, the process of calculating the work function is similar to that of a metal surface. However, semiconductors exhibit unique characteristics due to their energy band structure, which affects electron behavior.

## Solved Problems: How to Calculate the Work Function of a Metal

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

### Problem 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

### Problem 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

### Problem 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**

### Problem 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 0.2 eV

## More Videos: Work Function

**Work Function**

**Photoelectric Effect**

Drop a comment on what you don’t understand so that we can help you with more resources.

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How to Calculate Threshold Frequency

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**Reference**