When a light of wavelength \( \lambda \) falls on the emitter of a photocell, maximum speed of emitted photoelectrons is \( v \). If the incident wavelength is changed to \( \frac{2}{3} \lambda \), the maximum speed of emitted photoelectrons will be:
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The maximum speed of the emitted photoelectrons is proportional to the square root of the photon energy. When the wavelength is reduced, the energy increases, and so does the speed.
Step 1: Understanding the Question:
Photoelectric effect relating wavelength and maximum kinetic energy. We assume the work function is negligible compared to photon energy to compare velocity ratios. Step 2: Key Formula or Approach:
\[ K_{max} = \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \Phi \approx \frac{hc}{\lambda} \]
This implies \( v^2 \propto \frac{1}{\lambda} \) or \( v \propto \frac{1}{\sqrt{\lambda}} \). Step 3: Detailed Explanation:
Initially: wavelength \( \lambda \), speed \( V \).
Finally: wavelength \( \lambda' = \frac{2\lambda}{3} \), speed \( V' \).
Taking the ratio:
\[ \frac{V'}{V} = \sqrt{\frac{\lambda}{\lambda'}} = \sqrt{\frac{\lambda}{2\lambda / 3}} = \sqrt{\frac{3}{2}} \]
\[ V' = \sqrt{\frac{3}{2}} V \] Step 4: Final Answer:
The new maximum speed will be \( \sqrt{\frac{3}{2}} \) V.