Question

(1 point) A Silver surface (work function = 4.26 eV) is illuminated with 270 nm light from a 3 mW laser. How many photons per second are emitted by the laser? 4.0909E15 photons/s Only a small fraction of these photons can interact with electrons at the surface. Suppose this fractional efficiency is 0.00031%. What is the current in the detector assuming all the emitted electrons are captured? пA Note that current is charge(in Coulombs) per second. The charge of one electron is 1.6 x 10-''C

Answer 1

Silven surface coook function w = 4.2000 wavelength of incident light = 270mm Int power of light p= 3 mW. Now let 'N'No. of photons incident per unif time. energy of each photon E = ho Then NE = p perunit - Whe= 3x103 / N = 3x10 3 x 270 x109 6.6x1034 x 3x108 = 3x270 1714 6.643 - 40.9 41019 photone ons (plank conto N = 9.09% 10 photon, see time ha 6.6x20 39 Jos Its -9.09 x 10151 persee

electron emission efficiency n = 0.00031% So Emitted eleton Non incident photon No. Ne = 0.00031 1oo 100 OO 100 → Ne - Nx 0.00031 = 9.09x10' 0.0003|| Peletion / Ne = 12.67x10 9 persee I Number Then eletric current corresponding These elehousis I = (Ne) xq att utsee I=12.67x10x1,6x109 - I = 2.028 NA I

nm potassium surface stopping potential V = 0.58V correspond to x1 = 433 nm. and stopping potential V2 = 2.29 Volt- comsponds to 1 271nm. Then eve = he -W - 6 I 7 h = ? Vad w = 1.6x10tx 2.29x/271x10f 3x108 and L I = 10 di-W !! Then e (V2-vi) = he [1, 17 h = e(va-vi) C[ - ] and ev = he wto N

h= 1.6x10 x (2.29-0.58) 30 100 x 102 7410 19 - 1.68 1.71x10 A-141x1014 a h = 6.60 x10 39 J. 6) Now ev = he - w & wehe -evi -34 - 6.6x10 x 3 xH 433x 159 W J = 6.6x1034 x3x108 1.6x1019 8483x10 9 -0.58 TW = 2.27 ev /

o cut off wavelength Cao cwhoff, helo = w l wavellent no = he = 5.45x107m ho = 545 am