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EEJ MAIN Physics QUESTION #998
Question 1
A light source of wavelength \(\lambda\) illuminates a metal and electrons are ejected with maximum KE of 2 eV. If the same surface is illuminated by wavelength \(\dfrac{\lambda}{2}\), the maximum KE of ejected electrons will be (work function \(= 1\ \text{eV}\)):
  • 3 eV
  • 2 eV
  • 6 eV
  • 5 eV✔️
Correct Answer Explanation
From first case: \(\frac{hc}{\lambda} = \phi + 2 = 1+2 = 3\ \text{eV}\). With \(\frac{\lambda}{2}\): energy \(= \frac{2hc}{\lambda} = 6\ \text{eV}\). Max KE \(= 6 - 1 = 5\ \text{eV}\).