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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10.1.1 The work function is the minimum energy required to eject an electron from the surface of a metal. 10.1.2 Step 1: Identify the condition for threshold frequency. When the maximum kinetic energy of emitted electrons is 0 J, the incident frequency is equal to the threshold frequency. Step 2: Read the incident frequency from the problem statement. The incident light frequency is given as 1.045 × 10^15 Hz. Step 3: State the threshold frequency for Zinc. Since the maximum kinetic energy for Zinc is 0 J, its threshold frequency is equal to the incident frequency. f_0 = 1.045 × 10^15 Hz 10.1.3 Step 1: Recall the photoelectric effect equation. The energy of an incident photon (hf) is used to overcome the work function (W_0) and provide kinetic energy to the emitted electron (K_max). hf = W_0 + K_max Rearrange to find the work function: W_0 = hf - K_max Step 2: Calculate the energy of the incident photon. Given frequency f = 1.045 × 10^15 Hz and Planck's constant h = 6.626 × 10^-34 J·s. E = hf = (6.626 × 10^-34 J·s) × (1.045 × 10^15 Hz) E = 6.92417 × 10^-19 J Step 3: Calculate the work function for Sodium. From the table, for Sodium, K_max = 2.53 × 10^-19 J. W_0, Na = (6.92417 × 10^-19 J) - (2.53 × 10^-19 J) W_0, Na = 4.39417 × 10^-19 J Step 4: Calculate the work function for Caesium. From the table, for Caesium, K_max = 3.50 × 10^-19 J. W_0, Cs = (6.92417 × 10^-19 J) - (3.50 × 10^-19 J) W_0, Cs = 3.42417 × 10^-19 J Step 5: Compare the work functions and provide a reason. 4.39417 × 10^-19 J > 3.42417 × 10^-19 J. The work function of sodium is GREATER THAN that of caesium. Reason: Caesium has a higher maximum kinetic energy for the same incident light frequency, which means less energy was required to eject electrons, indicating a smaller work function. 10.1.4 NO Reason: Electron emission depends on the frequency of the incident light (the energy of individual photons). If the frequency is below the threshold frequency of the metal, no electrons will be ejected, regardless of how high the intensity (number of photons) is. For metal M, electrons are not emitted at the given frequency, meaning this frequency is below its threshold frequency. 10.2.1 Step 1: Determine the energy of the emitted photon. When an electron moves from a higher energy level (E_3) to a lower energy level (E_1), the energy of the emitted photon is the difference between these levels. E_3 = 5.7 × 10^-19 J E_1 = 1.3 × 10^-19 J E = E_3 - E_1 = (5.7 × 10^-19 J) - (1.3 × 10^-19 J) E = 4.4 × 10^-19 J Step 2: Calculate the frequency of the photon. Use the formula E = hf, where h = 6.626 × 10^-34 J·s. f = ( E)/(h) = 4.4 × 10^-19 J6.626 × 10^-34 J·s f = 6.639 × 10^14 Hz The frequency of photon X is 6.64 × 10^14 Hz (rounded to three significant figures). 10.2.2 NO Reason: An emitted photon's energy must correspond to the exact energy difference between two allowed energy levels in the atom. The possible energy differences for this atom are: • E_3 - E_2 = (5.7 - 5.1) × 10^-19 J = 0.6 × 10^-19 J • E_3 - E_1 = (5.7 - 1.3) × 10^-19 J = 4.4 × 10^-19 J • E_2 - E_1 = (5.1 - 1.3) × 10^-19 J = 3.8 × 10^-19 J Since 2.5 × 10^-19 J does not match any of these allowed energy differences, a photon with this energy cannot be emitted. Send me the next one 📸