Belongs to the modern physics in which we have a photo cathode that is emitting electrons with maximum kinetic energy E it is equals to 0.85 electron volt. When illuminated with wavelength lambda equals to 4 30 nanometer violet light. So for the part we have to determine village exact electron under equivalent lambda read. It is equal to 6 33 nanometer. Okay, so for this question, first of all, calculating the work function of the material. So we can write from the kinetic photoelectric effect equation so we can write that At Sea Belinda. This will be equals two kinetic energy maximum plus five. Okay, so from here we get five. This is equals two at symbol lambda minus kinetic energy. Okay, so maxim kinetic energy, it can be simply return its kinetic energy. So substituting values so we get five. It is equal to X, which is Planck's constant 6.626 molecular way 10 to the power -34, replaced by C. Is the speed of light. So 3 to 10 to the power eight divided by incident wavelength which is 4 30 nanometer. So 10 to the power minus nine minus kinetic energy which is this value. So 0.85 particular by 1.6 to 10 to the power minus 19. And this value comes out to be in jewel. So from here we get five. This is one function of the material, so it is equal to 3.26 molecular by 10 to the power minus 19 jewel. Okay, now 40 part A. We have this equivalent so calculating energy for this equivalent solar energy for the red light, it is equals two at Ciba lambda Red. So substituting value. So we get 6.626 molecular by 10 to the power minus 34 molecular by three and 2. 10 to the power eight m per second divided by lambda Red which is 6 33 particular by 10 to the power -9 m from here we get energy for the red light. It is equal to 3.14 particular by 10 to the power minus 19 jewel. So here we can see that we're function. Fight it is greater than the energy of incident light that is red light. So from the theory of photo electrons, we can see here that the photo electrons will not be omitted or we can say that there will be no ejection of photo flecktones when red light is used. So the answer for this question will be we can say that No. Okay, the answer will be here. No. The photo flecktones under red light will not be admitted or rejected. Now moving to the part B. So we have to calculate here the three short wavelength for this material that is linda. Not So linda, not, it is equal to at sea by five, So substituting their lives. So edges against 6.626 multiply by 10 to the power minus 34 june 2nd C. Is the speed of light, so 3 to 10 to the power eight m per second, divided by five, which is three point 26 molecular weight 10 to the power minus 19 jewels. So from here, three should be linda. Not. It is equal to 609.890 m. So this becomes the answer for the part B of the problem. Okay.