mss90 said:Ek=(hc/lambda)-W
((6.63*10^-34)*(3*10^8))/(2.15*10^-7) - (5.888*10^-9)
(9.25*10^-19) - ( (5.888*10^-9) = -5.887*10^-9 J / (1.6*10^-19) = -3.68*10^10eV
The photoelectric effect is the phenomenon where certain metals emit electrons when exposed to light. This occurs because the energy from the light is transferred to the electrons, allowing them to overcome the binding forces of the metal and escape.
The intensity of light does not affect the photoelectric effect. Instead, it is the frequency of the light that determines the energy of the electrons emitted. Higher frequencies of light have more energy and can cause more electrons to be emitted from the metal surface.
The work function is the minimum amount of energy required for an electron to escape the metal surface. It is different for each type of metal and is a crucial factor in determining the energy of the emitted electrons in the photoelectric effect.
The photoelectric effect provides evidence for the particle nature of light, as it shows that light is made up of tiny packets of energy called photons. The energy of the emitted electrons is directly proportional to the frequency of the light, which supports the idea that light is made up of discrete particles.
The photoelectric effect has many practical applications, including solar panels, photodiodes, and photomultiplier tubes. It is also used in many devices such as cameras, scanners, and photocells. Additionally, the photoelectric effect plays a crucial role in the development of quantum mechanics and our understanding of the behavior of light and matter.