The stopping potential for electrons ejected by 6.8 * 10¹⁴-Hz electromagnetic radiation incident on a certain sample is 1.8 V.The kinetic energy of the most energetic electrons ejected and the work function of the sample, respectively, are: A)1.8 eV, 2.8 eV B)1.8 eV, 1.0 eV C)1.8 eV, 4.6 eV D)2.8 eV, 1.0 eV E)1.0 eV, 4.6 eV

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Quizplus · Accepted Answer

The kinetic energy (KE) of the most energetic ejected electrons is equal to the stopping potential times the charge of an electron, which is 1.8 eV (since 1 eV = 1 V * charge of an electron). The work function (ϕ) can be calculated using the photoelectric equation: KE = hf - ϕ, where h is Planck's constant and f is the frequency of the incident light. Since we are not given the exact value of hf, but we know KE = 1.8 eV, we can infer that the work function is the difference between the photon energy (hf) and the kinetic energy (1.8 eV). Without the exact value of hf, we cannot directly calculate ϕ from the given information, but the correct answer must be consistent with the relationship KE + ϕ = hf. Given that the stopping potential is 1.8 V, and assuming the incident photon energy is greater than the work function, the only option that correctly reflects the relationship between the kinetic energy and the work function without needing the exact value of hf is option B, where the kinetic energy is 1.8 eV, and the work function is inferred to be the difference needed to match the photon energy, which is not directly calculable from the given information but is consistent with the options provided.

The Stopping Potential for Electrons Ejected by 6