Question

Question 7 1 pts When light of wavelength , illuminates the surface of Metal 1, the stopping voltage is V. In terms of V, what will be the stopping voltage if the same wavelength is used to illuminate the surface of Metal 2? The work function of Metal 1 is 5.7 eV and the work function of Metal 2 is 2.1 eV. 3.6 volts - V O V-3.6 volts OV+3.6 volts 2.1 volts - V 5.7 volts - V

Answer 1

From Einstein's photoelectric equation, we know that ev= nu-nuo hvor work function , hua incident enersy v- stopping potential. Stay Assume that stopping potential for and metal in v ev=ne - 5.7 ev- for metal to evane - 2.lev.- for metal subtracting. o. ver'= -5.7+2.1 =) V'V +36 so, the 3rd option is correct 2 Here, option 3 (11&N only) is correct Enersy of emitted photo electrons can't be greater than that of incident light => Energy will not be conserved -> Eliminater i Scattered light will have less enersy => longer. wavelength ( Energy conservation) => Eliminates on.in are correct => 11 & in principle, From Heisenberg uncertainty - 4x xap 2 h 12 -> ? 6.6*10-34 9:1716-344716721 =) 4x in 10-12 => Option 2 is correct