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In a photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be

1. (3/4)^(1/2) v
2. (4/3)^(1/2) v
3. less than (4/3)^(1/2) v
4. greater than (4/3)^(1/2) v

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

Solution

The energy of the emitted electron is related to the wavelength of the incident light by the photoelectric equation. Reducing the wavelength to 3λ/4 increases the photon energy, which increases the kinetic energy of the emitted electron. The speed of the electron is proportional to the square root of its kinetic energy, leading to the factor (4/3)^(1/2).

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