Question

***Please help fill in the table finding KE...& the following questions

Metal Wavelength (nm) Frequency (Hz) f Stopping Voltage (V) KE (J) Sodium 125 2.4*1015 -7.58 Sodium 300 1.0*1015 -1.79 Sodium 450 6.7*1014 -0.45 Sodium 538 5.57 1014 -0.01 Sodium 540 5.55*1014 0 a. Describe what happens to the stopping voltage for wavelengths greater than or equal to 540 nm. Based upon your knowledge of the atom, hypothesize an explanation for such behavior. b. At 540nm no electrons are ejected. The frequency or wavelength at which electrons are just barely ejected from the metal is referred to as the threshold frequency, fo. What the threshold frequency f, might be for the Sodium metal plate? c. According to atomic theory, electrons are bound to the nucleus of the atom because of the electrostatic attraction between electrons and the positive nucleus of the atom. If an electron is given enough energy, the electron will leave the atom, ionizing the atom. The binding energy, BE, for an atom is the minimum amount of energy needed to remove an electron from the atom. Based upon the data from the previous question, and using BE=hf, calculate the binding energy for Sodium in eV and in Joules./

Answer 1

Cosen 0.0IV Case ! V = – 7.58 volts K.E = ev 7.58 ev - 1.2128x10-18 J v=-1.79 volls K.E= 1.79 ev = 2.864810-19 Case? v=-0.45 volts K.E= 0.45 ev 7.2x10-20 J Case 4 v= K.E o.oiev K.E 1.6410-21 J 62 if d = 540 um, then the stopping potential energy (vollage) remains Zero. It is so be cause no electrons are ejected at lower energy (d>shoum) 6) Кс - so = 3x108 5.555x10' Hz (C) B.E = 6.63410344 s. 55sxolle 3.68341019 - 2.3 ev E = hf. d. 540x10-9