When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take 1.24 × 10⁻⁶ J·m·C⁻¹]

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In this photoelectric effect problem, we have a metal surface illuminated by light of a specific wavelength. The stopping potential needed to prevent photoelectrons from reaching the positive electrode is 6.0 volts. When a different light source with four times the wavelength and half the intensity is used, the stopping potential drops to 0.6 volts. We need to find the wavelength of the first source and the work function of the metal. We're given that the constant h times c divided by e equals 1.24 times 10 to the negative 6 joule-meter per coulomb. Let's set up the equations for this problem. The photoelectric effect equation states that the stopping potential times the elementary charge equals the energy of the incident photon minus the work function. For the first source with stopping potential 6.0 volts, we have e times 6.0 equals h times c divided by lambda 1 minus phi. For the second source with stopping potential 0.6 volts and wavelength 4 times lambda 1, we have e times 0.6 equals h times c divided by 4 lambda 1 minus phi. We're given that h times c divided by e equals 1.24 times 10 to the negative 6 joule-meter per coulomb. We'll use these equations to solve for lambda 1 and phi. Now let's solve the system of equations. First, we set up the equations using the photoelectric effect formula. For the first source, we have e times 6.0 equals h times c divided by lambda 1 minus phi. For the second source, we have e times 0.6 equals h times c divided by 4 lambda 1 minus phi. Next, we substitute A equals h times c divided by e, which is given as 1.24 times 10 to the negative 6 joule-meter per coulomb. We also let W equal phi divided by e to simplify our equations. This gives us 6.0 equals A divided by lambda 1 minus W, and 0.6 equals A divided by 4 lambda 1 minus W. Subtracting the second equation from the first, we get 5.4 equals A divided by lambda 1 minus A divided by 4 lambda 1, which simplifies to 5.4 equals 3A divided by 4 lambda 1. Now we'll solve for the wavelength and work function. From our previous equation, 5.4 equals 3A divided by 4 lambda 1, we can solve for A divided by lambda 1, which gives us 7.2 volts. Next, we substitute this value back into the equation 6.0 equals A divided by lambda 1 minus W to find W. This gives us W equals 1.2 electron volts, which is the work function in electron volts. To find lambda 1, we use the equation A divided by lambda 1 equals 7.2 volts, where A equals 1.24 times 10 to the negative 6 joule-meter per coulomb. Solving for lambda 1, we get 1.72 times 10 to the negative 7 meters, or 172 nanometers. Finally, to convert the work function from electron volts to joules, we multiply by the elementary charge: 1.2 electron volts times 1.602 times 10 to the negative 19 joules per electron volt, which gives us 1.92 times 10 to the negative 19 joules. So our final answers are: the wavelength of the first source is 172 nanometers, and the work function of the metal is 1.92 times 10 to the negative 19 joules. Let's summarize what we've learned. We've found that the wavelength of the first light source is 1.72 times 10 to the negative 7 meters, or 172 nanometers, which is in the ultraviolet range. The work function of the metal is 1.92 times 10 to the negative 19 joules, or 1.2 electron volts. This problem demonstrates key principles of the photoelectric effect. First, the energy of the incident photons depends on their frequency or wavelength, not their intensity. The shorter the wavelength, the higher the energy. Second, the maximum kinetic energy of the emitted photoelectrons equals the photon energy minus the work function. This is why the stopping potential is higher for the first source with its shorter wavelength. The energy diagram shows that the first source with 172 nanometer wavelength provides 7.2 electron volts of energy, resulting in photoelectrons with 6.0 electron volts of kinetic energy. The second source with 688 nanometer wavelength provides only 1.8 electron volts, resulting in photoelectrons with just 0.6 electron volts of kinetic energy. This confirms Einstein's explanation of the photoelectric effect, which was a crucial development in quantum physics.