Question

X9 Identify the phenomena that suggest EMR must be treated as a particle. photoelectric effect, Compton Effect, diffraction b. interference, photoelectric effect, diffraction c. photoelectric effect, Compton Effect diffraction, interference, refraction d. 10. Zinc has a work function of 40 eV and is used in a photoelectric cell with light of an energy of 19 eV per photon. Determine the maximum kinetic energy of photoelectrons leaving the cell a. 11 eV 6 15 eV c. 19 eV d. 23 eV 11. The intensity of light is manipulated in a photoelectric experiment. When its frequency is above the threshold frequency, the a. energy of each photoelectron will change b. maximum kinetic energy of the electrons will change c. number of photoelectrons will change d. frequency of the incident photons will change 12. Radiation with a frequency of 7.52 x 10 Hz is incident on a photoelectric surface. A A stopping voltage of 0.916 V is required to reduce the current through the cell to zero. The work function of the photoelecttic surface is a. 1.10 eV b. 4.80 x 10-19 J c. 1.47 x 10-19 W d. 2.20 eV
Questiond  9,11 and 12. how would you aplroach these

Answer 1

@ Answer: Explanation. In photo electric effect when a radiation of suitable frequency strikes a metal surface photo electrons emit. Electron emission takes place due to collision between photon and electron. As collision is a particle property, we can say EMR can be treated as particle (photons). » In compton effect also collision between photon and electron leads to decrease in energy of photon.

. Above there shold frequency if intensity of light is increased, then the number of photons that are incidenting on photometal increase which leads to increase in the number of photos elections. Photo current & Intensity of light source. Namber photo electrins persee con Photo current Intensity Given frequency (f) = 7.52X10" Hz 8: 916 V. stopping potential (Vo) = To find work function, we can use Einstien's photo electric equation. kar je bf = Kmaxt ¢ Kmax = evo = maximum Kinetic energy of photoelectrons. d= work function. -&- hf. Kmax 30 Kmax = evo = 1.6x10" x 0.916 = 1.4656 x 1019 J hf = 6.625x10x 7.52x10". 49-82 x 1080 hf= 4.982x109 T

using the values in eq 0 we get, 6= 4.982x10' ? 1.4656x15 19 -19 = 3.5164 21095 = &= 3.5164 X18 1.6x589 2-2 év $ 2.1970V