laser eye surgery after detached retina

quantum physics help please?

1. After applying sunscreen, Cherie is in the summer sun to get a tan. The UV light responsible for tanning has a wavelength of 310 nm, while the burning rays of up to 280 nm. Ultraviolet photons emitted more energy? How much energy in Joules? 2.Mitch to undergo eye surgery to repair a detached retina. Your doctor uses a laser color Green whose wavelength is 541 nm. How much energy is produced by each photon laser? Put your frequency response J 3.Light of 1.00 x 1015 Hz illuminates a sodium surface. The photoelectrons ejected are found to have a maximum kinetic energy of 1.86 eV. Calculate the work function of sodium. 4.Shelley shines its violet laser, helium-neon, whose wavelength is 400 nm in a photoelectric cell with an work function of 2.38eV. What is the kinetic energy of photoelectrons released? (In eV) have worked on these problems for an hour and I still can not understand. help would be greatly appreciated.

1) Thanks to some genius at the beginning of 20, we know that the quantity energy of a photon is directly proportional to its frequency. As the frequency is inversely proportional to the wavelength, the more short wavelength photons occur more frequently (and therefore higher energy). Proof: E0 = hf, where E0 is the energy of a photon, h is Planck's constant and f is the frequency of the photon. E0 = hf = h (w / w) = H c / w, where c is the speed of light, w is the length wave (derived from the equation c = wf, part of the wave theory of the standard). E0 = (6.626e-34) (3E8) / (3.1e-7 m) = 6.41e-19 J = E0 (6.626e-34) (3E8) / (2.8E-7 m) = 7.10e-19 J So you can see the wavelength of 280 nm photons have higher energy. 2) Using the formula in the last question: E0 = hc / w = (6.626e-34) (3E8) / (5.41e-7m) = 3.67e-19 J 3) To do this, the kinetic energy acquired by an electron by absorption of a photon is the photon energy, less is known than the work function (which is the amount of energy required to release electrons from their atoms). W = Ek W = W-E0 E0-Ek-Ek = hf = (4.14e-15) (1.00e15) – (1.86) = 2.28 eV is the last unit not indicated in question, I guess it's OK to leave the electronic response. 4) You can use the same equation in question 3 above. Ek W-= E0 = hc / ww = (4.14e-15) (3E8) / 4.0E-7 m) = -2.38 0.725eV And there are all your answers. Note: If you're wondering why I used two values different Planck's constant, is to take into account the change of units. EVs are in joules, which measure the amount of energy. An electron-volt is the amount of energy an electron gains when placed in a potential difference of 1V. Therefore E = QV = (1.60e-19) (1) = 1.60e-19 J. Thus we see that 1.60e-19J is an energy eV. Planck's constant is 6.626Js. To turn what can be used with electrons, it Just use the conversion factors. 6.626Js * (1eV/1.60e-19J) = 4.14125e-15 EV = 4.12e-15 EVS (2 decimal places is fine for these issues. This is done simply reduce the amount of work and allow us to obtain eV easy answers, that the question relates. Greetings.