In photoelectric emission process from a metal of work function $1.8\, eV$, the kinetic energy of most energetic electrons is $0.5\, eV$. The corresponding stopping potential is
- 1.8V
- 1.3V
- 0.5 V
- 2.3 V
The Correct Option is C
Approach Solution - 1
$V _{0}=\frac{ K \cdot E _{\max }}{ e }=\frac{0.5 \,eV }{ e }$
$V _{0}=0.5$ volt
Approach Solution -2
The stopping potential Vs is related to the maximum kinetic energy of the emitted electrons Kmax = e Vs Kmax =0.5 eV or Vs = 0.5 V
Therefore, the correct option is ‘C’ i.e 0.5V
Approach Solution -3
In a photoelectric effect, the total maximum kinetic energy of an electron is,
Kmax= eV0
So, V0= \(\frac{k_{max}}{e}\) = \(\frac{0.5eV}{e}\) = 0.5V (as eV = 1 electronic charge time at 1 Volt)
Therefore, V0= 0.5V
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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.