Solution:
The maximum kinetic energy of emitted photoelectrons is related to the stopping potential by \( K_{\max} = e V_s \). Given \( V_s = 2.5 \text{ V} \), the maximum kinetic energy is simply \( 2.5 \text{ eV} \).
In an experiment on photoelectric emission for incident light of wavelength \( 1.98 \times 10^{-7} \text{ m} \), stopping potential is found to be \( 2.5 \text{ V} \). What is maximum kinetic energy of emitted photoelectron?
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