The work function for metallic rubidium is 2.09 eV. Calculate the kinetic energy and the speed of the electron ejected by light of wavelength (i) 650 nm and (ii) 195 nm.
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Answer:
Strategy: This problem is related to the photoelectric effect. Work function is the minimum amount of energy needed to remove an electron from metal surface. Light radiation has energy equal to more than the work function can eject the electron metal surface. Some enrgy of the radiation is taken to eject the electon and rest is equal to the kinetic energy of the electron. In this problem (a) we have to calculate the kinetic energy of the elctron for Caesium and for that we are given two values of wavelengths. we can use that to get energy of each of them and use to get kinetic energy. Before doing calcuation we will bring all the units in the eV b) Here we have to get the electrons kinetic energy in Rubidum. Solution: a) Energy of photon = Work function+ K.E of electron hv = W + Kmax , where K max is the kinetic energy of the elctron., h is the planks constant , W is the work function. Kmax = hv-W, v is the frequency. h = 6.626E-34 J s We have to get frequency by using wavelenght. Since v=c/λ , where c is the speed of light and its value is 3.0E8 m/s Lets calculate hv For λ = 700 nm hv = hc/ λ = (6.626E-34 J *3.0E8m/s )/( 700 E-9 m ) , ……………………(here we used conversion of 1 nm = 10^-9 m, to get quantities in same unit. 2.8397E-19J This value in J should be converted to eV 1eV= 1.6E-19 J Energy value in eV = 2.8397E-19 J * 1 eV/1.6E-19 J =1.77482 eV This value is not more than or equal to the work...
Answer:
(a) The work function of metallic caesium is 2.14 eV. Calculate the kinetic energy and the speed of the electrons ejected by light of wavelength (i) 700 nm, (ii) 300 nm. (b) The work function of metallic rubidium is 2.09 eV. Calculate the kinetic energy and the speed of the electrons ejected by light of wavelength (i) 650 nm, (ii) 195 nm.
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