Question

Problem #4. An E&M wave having wave length of 560. nm hits a surface. Electrons were released without Kinetic energy (cut-off) i.e. this light's energy was just enough to release the electrons without speed. nm IF ANOTHER wave was used to the same surface having wave length of 400. Nm. A. What is the work function of this surface in ev? B. What is maximum kinetic energy of the emitted electrons in eV AND Joules as well as its speed? C. What is the De Broglie wave length of the emitted electrons? 7=560nm us 77 2 -400nm (им,

Answer 1

(A) energy required to release the electrons without speed • Work Function of the metal sweeface Work function W = hc do where h = 6.67X10 34 J-S C = 3x108 m/s d. bauelength ght required = 560nda W = 6.67X10 -34 Js X 3X100 mg 560nm . ☺ -26 W= 20.01 X10 J-m 560 Xión m W = 0.03573 X10 -17 J W -19 3.573 XO"J Iev. = 1.6 X6-19 J W = 3.573 81019 146 Ale!? W = 2.233 eV

energy by light of wavelength carried id=400nm E = hc 6.67XO 400 X/6-9 m JS X 3x10 m/s E 0.0500 X 10 J 5.000 1019 J E 5 - x lo ev 106X 10-19 E = 3•125 eV (B) Energy E = Wt Kinetic energy In Joule Kinetic ene E-W energy KE = 5.00 X 101J 5.00X10" I-3.573870'I KE = 10427 X10791 In eve KE - E-W KE = 3.125 eV KEK 0.892 el 21233 OV KE 0.89 eV 1933

는 speed (say) m= mass of electron of electrons KE : Imuz 2 02 2KE m 2 (KE) Al m 2 X 1,427 X la F31 وا -19 J 시 9.1876-81 kg 40234368/ 6.3136x102 m²/2 دوا 0056 xlo X/of mis (c) de Broglie lie wavelengths A ho mu 34 IS dia 6.67 X10 9.1870–31 Kg x 0:56X/04 m/s 9 ชช) d = 1.3089 x 10 d 1.309 nm
Summary: the kinetic energy of emmited electrons is the difference of energy of photons falling on the surface and the work function of surface.