Light of wave lenght 600 nm falls of metal surface bigt 0 work function of metal surfare 2.1 e.v. find K.E. of emited elecerention
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The photoelectric effect occurs when light hits a metal surface and electrons are emitted. Light consists of particles called photons, each carrying energy given by E equals h f, where h is Planck's constant and f is the frequency. When a photon strikes an electron in the metal, it can transfer its energy to the electron, potentially giving it enough energy to escape from the metal surface.
Einstein's photoelectric equation describes the energy balance in photoelectric emission. The equation is K E equals h f minus phi. Here, K E is the kinetic energy of the emitted electron, h f is the energy of the incident photon, and phi is the work function of the metal. The work function represents the minimum energy needed to remove an electron from the metal surface. When a photon hits an electron, its energy is divided between overcoming the work function and providing kinetic energy to the electron.
Let's organize the given information for our photoelectric effect problem. We have a wavelength of 600 nanometers and a work function of 2.1 electron volts. We need to find the kinetic energy of emitted electrons. The key relationships we'll use are c equals lambda f, relating wave speed to wavelength and frequency, and E equals h f for photon energy. We also need fundamental constants: Planck's constant h equals 6.626 times 10 to the minus 34 joule seconds, the speed of light c equals 3 times 10 to the 8 meters per second, and the conversion factor 1 electron volt equals 1.602 times 10 to the minus 19 joules.
Now let's calculate the photon energy step by step. First, we convert the wavelength from nanometers to meters: 600 nanometers equals 600 times 10 to the minus 9 meters. Next, we calculate the frequency using f equals c over lambda. Substituting our values: f equals 3 times 10 to the 8 divided by 600 times 10 to the minus 9, which gives us 5.0 times 10 to the 14 hertz. Then we calculate the photon energy using E equals h f. This gives us 6.626 times 10 to the minus 34 times 5.0 times 10 to the 14, equals 3.313 times 10 to the minus 19 joules. Finally, we convert to electron volts by dividing by 1.602 times 10 to the minus 19, giving us 2.07 electron volts.
Now we apply Einstein's photoelectric equation to find the final answer. K E equals E photon minus phi. Substituting our calculated values: K E equals 2.07 electron volts minus 2.1 electron volts, which gives us negative 0.03 electron volts. Since the kinetic energy is negative, this means no electrons will be emitted from the metal surface. The physical explanation is that the photon energy of 2.07 electron volts is less than the work function of 2.1 electron volts. Therefore, the incident light does not have enough energy to overcome the work function barrier, and photoelectric emission cannot occur in this case.