A light source of wavelength \( \lambda \) illuminates a metal surface, and electrons are ejected with a maximum kinetic energy of 2 eV. If the same surface is illuminated by a light source of wavelength \( \frac{\lambda}{2} \), then the maximum kinetic energy of ejected electrons will be (The work function of the metal is 1 eV).
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In the photoelectric effect:
- The photon energy is inversely proportional to the wavelength.
- Doubling the frequency (or halving the wavelength) doubles the photon energy.
- The kinetic energy of emitted electrons is given by \( K_{\max} = h\nu - \phi \).
Step 1: Utilize the photoelectric equation, which is expressed as \( K_{\max} = hu - \phi \). Here, \( K_{\max} \) represents the maximum kinetic energy of the emitted electrons, \( hu \) denotes the photon energy, and \( \phi \) is the work function of the metal.
Step 2: Calculate the photon energy corresponding to the initial wavelength \( \lambda \). Given that the initial kinetic energy is 2 eV, we have \( hu = K_{\max} + \phi = 2 + 1 = 3 \text{ eV} \). Since \( hu = \frac{hc}{\lambda} \), it follows that \( \frac{hc}{\lambda} = 3 \text{ eV} \).
Step 3: Determine the new kinetic energy for a wavelength of \( \lambda/2 \). The photon energy for \( \lambda/2 \) is \( hu' = \frac{hc}{\lambda/2} = 2 \times \frac{hc}{\lambda} = 2 \times 3 = 6 \text{ eV} \). Consequently, the new kinetic energy is \( K_{\max}' = 6 - 1 = 5 \text{ eV} \). The final answer is \( \boxed{5} \).
