Question

!) Light with a wavelength of (465 + A) nm is incident on a metal surface with a work function of (1.25 + (0.1)(B)) eV. Find the maximum kinetic energy of the emitted photoelectrons. Give your answer in eV and with 3 significant figures.

2) Consider electrons accelerated using a potential difference of (12.5 + A + B) kV before hitting a metal surface. Calculate the minimum wavelength of the emitted x-rays. Give your answer in pm and with 3 significant figures.

A: 1

B: 4

Answer 1

1) A = 1 and B = 4 Wanelength of incident light = = (465 tanm t=1465+1) nm =) =466 nm speed of light =c=3x108 ms he E = of ㅢ where -15 n is blan ik's Constant 4.136 X10 eus is So, energy of incident light photon 4.136 x 10 els X3x10 ms' -15 E = 466 nm -7 =) E = [:enm = 10-9m m. 1 2. 4o8 x 10 evm 466 X 10 9 -7 12.408 X I eum 4.66 x 10-7 m = E - = E=2.662 66le v - W Work function of metal surface W = 11:25 + (0:1)(B) er => W (1.25 + (0-1) (")) ev W = 1.65 ev (1-25 +0.4) er Let maximum electrons KE max Kinetic energy of emitted photo- = photo electric effect, we = Now, By By Einstein's equation of have Work function of metal Energy of Incident light + maximum Kinetic energy of photoelectron E =w +KEmax =) Kemax 2.6626612V - 1.6 5ev =) Kemax 1001266 eV Ž 100 lev. So, maximum Kinetic is Kemax = E -W - Х energy of photo electron 1.olev