Answer Happy

B = B= = Atobserver yAtmoving , Lobserver L moving 7 C=3.0 x 108 m/sec, V1 - B2 V1 + V2 B| А = UBIC + UCLA , W = 1+ U102 E = hf hc , df = c, KEmax = eV, = hf – 0, X - to = h (1 mec cose), 1 EO ph m , c = 3.0 x 108 m/sec , me = 9.11 x 10-31 kg, Eph + KE, KE zmev?, 1 nm = 1 x 10 h = 4.14 x 10-15 eV · sec , h = 6.63 x 10 2 -34 J · sec 1 eV = 1.60 x 107 -19 J, 1 keV - 1000 eV 7
Problem 3: When light with a wavelength of 450.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 0.26 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 200.0 nm falls on the same surface?