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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Right Maths, let's go. 10.1 The work function of a metal is the minimum energy required to eject an electron from the surface of the metal. 10.2 Step 1: Determine the work function from the graph. From the graph, the threshold energy (the energy of incident photons at which v_max^2 = 0) is the work function (W_0). W_0 = 5.4 × 10^-19 J Step 2: Compare the calculated work function with the values in the table. The table shows: • Barium: 4 × 10^-19 J • Chromium: 5 × 10^-19 J • Aluminium: 7 × 10^-19 J The calculated work function 5.4 × 10^-19 J does not exactly match any of the given values. However, the graph shows the x-intercept at 5.4 × 10^-19 J, which represents the work function. There might be a slight discrepancy or rounding in the table values, or the question expects us to use the graph's intercept as the exact work function for the experiment. Given the options, none of the metals perfectly match 5.4 × 10^-19 J. Let's re-examine the graph and question. The question asks to "identify the metal used". The graph is for this experiment. The x-intercept is the work function for this experiment. The table lists possible metals. It's possible the value 5.4 × 10^-19 J is the actual work function of the metal used, and the table is just for comparison, or there's an intended closest match. Let's assume the question implies that one of the metals in the table was used, and we need to find the closest match or if there's a misinterpretation. The work function is the energy required to eject an electron. From the graph, this is the x-intercept where v_max^2 = 0. The value is 5.4 × 10^-19 J. None of the metals in the table have a work function of exactly 5.4 × 10^-19 J. However, the question states "One of the metals in the table below was used in this experiment." This implies that the work function derived from the graph must correspond to one of the metals. Let's re-read the graph carefully. The x-axis is "Energy of incident photons (x \ 10^-19 J)". The intercept is clearly at 5.4. The table values are also in (x \ 10^-19 J). Barium: 4 Chromium: 5 Aluminium: 7 There seems to be a mismatch between the graph's intercept and the table values if we are to pick one exactly. However, in physics problems, sometimes values are approximate or there's a closest match. If we consider the graph's intercept as the experimental work function, and we must choose from the table, there might be an error in the question or the table. Let's assume the graph is accurate for the experiment. The work function is 5.4 × 10^-19 J. If we must choose from the table, Chromium (5 × 10^-19 J) is the closest. But 5.4 is significantly different from 5. Let's consider the possibility that the graph's intercept is not exactly 5.4, but rather 5. It's hard to tell precisely from the image. But it's marked as 5.4. Let's proceed with W_0 = 5.4 × 10^-19 J as derived from the graph. If the question insists on choosing from the table, and there's no exact match, it's problematic. However, the question asks to "Use a suitable calculation to identify the metal used." The calculation is finding W_0 from the graph. If the graph is for this experiment, then the metal used has a work function of 5.4 × 10^-19 J. Given the options, and assuming there might be a slight graphical interpretation or rounding, Chromium (5 × 10^-19 J) is the closest value to 5.4 × 10^-19 J. But this is an assumption. Let's consider the possibility that the question implies the work function is one of the table values, and the graph is a representation that might not be perfectly to scale for the intercept. However, the most direct interpretation is that the graph shows the work function. Let's assume the value 5.4 × 10^-19 J is the work function of the metal used. Since we must choose from the table, and 5.4 is not in the table, there might be an issue with the question's data. If I have to pick the closest one, it would be Chromium. But this is not a "suitable calculation" to identify it, it's an approximation. Let's re-evaluate the photoelectric equation: K_max = E - W_0. (1)/(2)mv_max^2 = E - W_0. When v_max^2 = 0, then E = W_0. From the graph, this occurs at E = 5.4 × 10^-19 J. So, the work function of the metal used in this experiment is 5.4 × 10^-19 J. Since the question asks to identify the metal used, and states "One of the metals in the table below was used", there must be a match. Given the options, and the clear marking of 5.4 on the graph, it's highly probable that the intended answer is Chromium, assuming some tolerance or a slight inaccuracy in the graph's exact intercept value or the table's values. If we strictly follow the graph, the metal has a work function of 5.4 × 10^-19 J. If we must choose from the table, Chromium is the closest. Let's assume the question expects us to use the graph's intercept as the work function for the metal used in the experiment. Work function W_0 = 5.4 × 10^-19 J. Comparing this to the table, Chromium has a work function of 5 × 10^-19 J. This is the closest value. The metal used is Chromium. 10.3 Step 1: State the photoelectric equation. The photoelectric effect equation is K_max = E - W_0. We also know K_max = (1)/(2)mv_max^2. So, (1)/(2)mv_max^2 = E - W_0. Step 2: Substitute the known values. From 10.2, the work function W_0 = 5.4 × 10^-19 J. The energy of incident photons for X is E = 7 × 10^-19 J. The mass of an electron m = 9.11 × 10^-31 kg. We need to calculate v_max^2, which is X. Step 3: Calculate K_max. K_max = (7 × 10^-19 J) - (5.4 × 10^-19 J) K_max = 1.6 × 10^-19 J Step 4: Calculate v_max^2 (X). (1)/(2)mv_max^2 = K_max v_max^2 = 2K_maxm v_max^2 = 2 × (1.6 × 10^-19 J)9.11 × 10^-31 kg v_max^2 = 3.2 × 10^-199.11 × 10^-31 m^2s^-2 v_max^2 ≈ 3.5126 × 10^11 m^2s^-2 The y-axis is labeled v_max^2 \ (x \ 10^11 m^2s^-2). So, X = 3.5126. Rounding to two decimal places as per the graph's precision for 6.59: X ≈ 3.51 The value of X is 3.51 10.4 The maximum velocities of the ejected electrons REMAINS THE SAME. Reason: The maximum kinetic energy (and thus maximum velocity) of the ejected electrons depends on the energy of the individual incident photons and the work function of the metal (K_max = E - W_0). Increasing the intensity of the incident light means increasing the number of photons, not the energy of each individual photon. Therefore, more electrons will be ejected, but their maximum kinetic energy and velocity will not change. Send me the next one 📸