The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14eV, 2.30 eV and 2.75eV respectively. If incident electromagnetic radiation has an incident energy of 2.20eV, which of these photosensitive surfaces may emit photoelectrons?
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14eV, 2.30 eV and 2.75eV respectively. If incident electromagnetic radiation has an incident energy of 2.20eV, which of these photosensitive surfaces may emit photoelectrons?
Na only
Cs only
Both Na and K
K only
The Correct Option is B
Approach Solution - 1
The correct option is (B): The minimum energy required for a photoelectron to escape from a metal surface is given by its work function. If the energy of the incident photon is greater than the work function of the metal, then photoelectrons will be emitted.
For the given incident energy of 2.20eV, only Caesium (Cs) can emit photoelectrons because its work function (2.14 eV) is less than the incident energy. The work functions of Potassium (K) and Sodium (Na) are greater than the incident energy, so they cannot emit photoelectrons under these conditions.
Approach Solution -2
Given energy of photon E=2.20eV
Work function of CS \(\phi_0=2.14eV,K\phi_0=2.30eV,Na \phi_0=2.75eV\)
We know that e- emitts when h\(\nu\)>\(\phi_0\)
So the energy of a photon is more than the work function of Cs.
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Concepts Used:
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Electronic Emission
The least energy that is needed to emit an electron from the surface of a metal can be supplied to the loose electrons.
Photoelectric Effect
The photoelectric effect is a phenomenon that involves electrons getting away from the surface of materials.
Heisenberg’s Uncertainty Principle
Heisenberg’s Uncertainty Principle states that both the momentum and position of a particle cannot be determined simultaneously.