This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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10.1 Define the term work function of a metal. The work function () of a metal is the minimum amount of energy required to eject an electron from the surface of that metal. 10.2 Use a suitable calculation to identify the metal used. Step 1: Identify the work function from the graph. The graph shows v^2_max versus the energy of incident photons. According to the photoelectric effect equation, E_k_max = E_photon - . When v^2_max = 0, the kinetic energy is zero, meaning E_photon = . This point is the x-intercept of the graph. From the graph, the line intersects the x-axis at 4 × 10^-19 J. Therefore, the work function of the metal is = 4 × 10^-19 J. Step 2: Compare the work function to the given table. • Barium: 4 × 10^-19 J • Chromium: 5 × 10^-19 J • Aluminium: 7 × 10^-19 J The metal used in the experiment is Barium. 10.3 Calculate X. Step 1: State the photoelectric effect equation relating kinetic energy to photon energy and work function. The maximum kinetic energy of an ejected electron is given by: E_k_max = E_photon - We also know that E_k_max = (1)/(2)mv^2_max. Substituting this into the equation: (1)/(2)mv^2_max = E_photon - Rearranging for v^2_max: v^2_max = (2)/(m)(E_photon - ) Where m is the mass of an electron (9.11 × 10^-31 kg) and is the work function (4 × 10^-19 J from 10.2). Step 2: Substitute the given values to calculate X. X represents v^2_max when E_photon = 5.4 × 10^-19 J. X = (2)/(9.11 × 10^-31) kg(5.4 × 10^-19 J - 4 × 10^-19 J) X = (2)/(9.11 × 10^-31)(1.4 × 10^-19) X = (2.19539 × 10^30) × (1.4 × 10^-19) X = 3.0735 × 10^11 m^2 · s^-2 The y-axis label is v^2_max (× 10^11 m^2 · s^-2), so X is the numerical coefficient. X ≈ 3.07 10.4 How will the maximum velocities of the ejected electrons change when the intensity of the incident light is increased? Choose from INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. The maximum velocities of the ejected electrons REMAINS THE SAME. Reason: The maximum kinetic energy (and thus maximum velocity) of photoelectrons depends only on the frequency (or energy) of the incident photons and the work function of the metal. Increasing the intensity of the incident light only increases the number of photons striking the surface per unit time, which in turn increases the number of photoelectrons emitted, but it does not change the energy of individual photons or the maximum kinetic energy of each ejected electron. Send me the next one 📸