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:
Dual Nature of Radiation and Matter
The dual nature of matter and the dual nature of radiation were throughgoing concepts of physics. At the beginning of the 20th century, scientists untangled one of the best-kept secrets of nature – the wave-particle duplexity or the dual nature of matter and radiation.
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.