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Monochromatic light of wavelength λ falls on a photocell, and the most energetic photoelectron is found to move with speed v. If the incident wavelength is reduced to 3λ/4, the speed of the fastest emitted electron will be:

1. v(4/3)^(1/2)
2. v(3/2)^(1/2)
3. > v(4/3)^(1/2)
4. < v(4/3)^(1/2)

Correct answer: > v(4/3)^(1/2)

Solution

Reducing the wavelength of the incident light increases the energy of the photons, which in turn increases the kinetic energy of the emitted photoelectrons. Since kinetic energy is proportional to the square of the speed, the fastest emitted electron will have a speed greater than v(4/3)^(1/2) when the wavelength is decreased to 3λ/4.

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