krit.club logo

Dual Nature of Radiation and Matter - Electron Emission

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Free electrons in a metal are held inside the metal by the attractive forces of the ions. To come out of the metal surface, an electron must gain a certain minimum amount of energy.

•

The minimum energy required by an electron to escape from a metal surface is called the Work Function of the metal, denoted by ϕ0\phi_0.

•

The work function depends on the nature of the metal and its surface properties. It is measured in electron volts (eVeV).

•

One electron volt (1eV1 eV) is the energy gained by an electron when it has been accelerated by a potential difference of 1volt1 volt.

•

Thermionic Emission: By heating the metal, sufficient thermal energy is imparted to the free electrons to enable them to come out of the metal.

•

Field Emission: By applying a very strong electric field (of the order of 108Vm−110^8 V m^{-1}) to a metal, electrons can be pulled out of the surface.

•

Photoelectric Emission: When light of suitable frequency illuminates a metal surface, electrons are emitted from the metal surface. These photo-generated electrons are called photoelectrons.

📐Formulae

1eV=1.602×10−19J1 eV = 1.602 \times 10^{-19} J

ϕ0=hν0\phi_0 = h \nu_0

ϕ0=hcλ0\phi_0 = \frac{hc}{\lambda_0}

💡Examples

Problem 1:

The work function of platinum is 5.65eV5.65 eV. Calculate its value in Joules (JJ).

Solution:

Energy(J)=5.65×1.602×10−19Energy (J) = 5.65 \times 1.602 \times 10^{-19} Energy=9.05×10−19J\text{Energy} = 9.05 \times 10^{-19} J

Explanation:

To convert energy from electron volts (eVeV) to Joules (JJ), we multiply the value by the electronic charge e=1.602×10−19Ce = 1.602 \times 10^{-19} C.

Problem 2:

If the work function of a metal is ϕ0=2.14eV\phi_0 = 2.14 eV, and the energy of an incident photon is 5.0×10−19J5.0 \times 10^{-19} J, will electron emission occur?

Solution:

ϕ0 in Joules=2.14×1.602×10−19J\phi_0 \text{ in Joules} = 2.14 \times 1.602 \times 10^{-19} J ϕ0≈3.43×10−19J\phi_0 \approx 3.43 \times 10^{-19} J Since 5.0×10−19J>3.43×10−19J5.0 \times 10^{-19} J > 3.43 \times 10^{-19} J, emission will occur.

Explanation:

Electron emission occurs only when the energy of the incident radiation is greater than or equal to the work function (ϕ0\phi_0) of the metal.