Maximum kinetic energy of photoelectrons – Rankers Physics
Topic: Modern Physics
Subtopic: Photoelectric Effects and deBroglie Equation

Maximum kinetic energy of photoelectrons

The stopping potential in the photoelectric experiment is \( 1.6\text{ V} \). The maximum kinetic energy of photoelectrons emitted is
\( 2.4 \times 10^{-19}\text{ J} \)
\( 2.56 \times 10^{-19}\text{ J} \)
\( 1.86 \times 10^{-19}\text{ J} \)
\( 1.4 \times 10^{-19}\text{ J} \)

Solution:

The maximum kinetic energy of photoelectrons is given by \( K_{\text{max}} = e V_0 \). Substituting the values: \( K_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 1.6\text{ V} = 2.56 \times 10^{-19}\text{ J} \).

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