I am having difficulty with this and have no idea where to start. What wavelengt
ID: 1745351 • Letter: I
Question
I am having difficulty with this and have no idea where to start.What wavelength of light must be incident upon a silver metalsurgace to cause an electron to have a kinetic energy equal to onehalf of the incident photon energy? If this wavelenth wereused in Einstein's apparatus, what stopping potential would benecessary to hold the electrons in place?
Any help would be appreciated greatly!
What wavelength of light must be incident upon a silver metalsurgace to cause an electron to have a kinetic energy equal to onehalf of the incident photon energy? If this wavelenth wereused in Einstein's apparatus, what stopping potential would benecessary to hold the electrons in place?
Any help would be appreciated greatly!
Explanation / Answer
We know the work function (Wo) of silver surface =4.73 eV = (4.73eV)(1.6*10-19J/eV) = 7.568*10-19J Given that kinetic energy of the emitted photoelectron isequal to half of the incident energy. Let be the wavelength of the incidnet light. Then Energy of the incident light rays is E = hc / Then kinetic energy of the emitted electron is Kmax = (1/2)(hc / ) We know the Einstein's photoelectric equation is E = Wo + Kmax hc / = Wo + (1/2)(hc / ) hc / - (1/2)(hc / ) = Wo (1/2)(hc/ ) = Wo (hc) / (2) = Wo = (hc) / (2Wo) ................(1) Here h = Planck's constant = 6.63*10-34J.s c =3.0*108m/s Wo =7.568*10-19J Substitute these values in the eq (1) we get the value of thewavelength () of the incident light. Now kinetic energy is Kmax = eVo (1/2)(hc / ) = eVo (hc) / (2) = eVo Vo = (hc) / (2e) ................... (2) Here h = Planck's constant = 6.63*10-34 J.s c =3.0*108m/s e =1.6*10-19C issubstituted here which is calculated from the eq (1).Do all the calculation work. (1/2)(hc/ ) = Wo (hc) / (2) = Wo = (hc) / (2Wo) ................(1) Here h = Planck's constant = 6.63*10-34J.s c =3.0*108m/s Wo =7.568*10-19J Substitute these values in the eq (1) we get the value of thewavelength () of the incident light. Now kinetic energy is Kmax = eVo (1/2)(hc / ) = eVo (hc) / (2) = eVo Vo = (hc) / (2e) ................... (2) Here h = Planck's constant = 6.63*10-34 J.s c =3.0*108m/s e =1.6*10-19C issubstituted here which is calculated from the eq (1).
Do all the calculation work. (hc) / (2) = Wo = (hc) / (2Wo) ................(1) Here h = Planck's constant = 6.63*10-34J.s c =3.0*108m/s Wo =7.568*10-19J Substitute these values in the eq (1) we get the value of thewavelength () of the incident light. Now kinetic energy is Kmax = eVo (1/2)(hc / ) = eVo (hc) / (2) = eVo Vo = (hc) / (2e) ................... (2) Here h = Planck's constant = 6.63*10-34 J.s c =3.0*108m/s e =1.6*10-19C issubstituted here which is calculated from the eq (1).
Do all the calculation work. (1/2)(hc / ) = eVo (hc) / (2) = eVo Vo = (hc) / (2e) ................... (2) Here h = Planck's constant = 6.63*10-34 J.s c =3.0*108m/s e =1.6*10-19C issubstituted here which is calculated from the eq (1).
Do all the calculation work. c =3.0*108m/s e =1.6*10-19C issubstituted here which is calculated from the eq (1).
Do all the calculation work.
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