Final answer:
In a photoemissive cell, when the exciting wavelength is changed to 3/4, the speed of the fastest emitted electron would typically increase to 4/3 times its original speed, given that photon energy is directly related to frequency and inversely related to wavelength.
Step-by-step explanation:
The question asks about the change in speed of the fastest electron emitted from a photoemissive cell, also known as the photoelectric effect, when the exciting wavelength is changed. According to the photoelectric effect, the energy of an incident photon is used to release an electron from the metal surface. The kinetic energy (KE) of the emitted electron is given by the equation KE = h*f - ϕ, where h is Planck's constant, f is the frequency of the incident light, and ϕ (phi) is the work function of the metal. Since f is inversely proportional to the wavelength (λ), a smaller wavelength means higher frequency, hence greater photon energy, leading to a faster emitted electron.
When the wavelength is changed to 3/4 of its original value, the frequency becomes 4/3 of the original frequency (since c = fλ where c is the speed of light). If we assume that the incident photon energy is just above the work function (ϕ), making the kinetic energy of the emitted electron directly proportional to the frequency, then typically, increasing the frequency by a factor of 4/3 will also increase the kinetic energy - and consequently the speed - of the emitted electron by the same factor. Therefore, the speed of the fastest emitted electron will be 4/3v.
However, since the question does not provide information about the work function or whether the initial photon energy is close to the work function, the final answer might need to consider these additional factors in a real-world scenario.