the work function of a metal is 4.2eV. If the radiation of 2000 angstrom falls on the metal, find the kinetic energy of the metal.
Hello Adiba Ruquiya Begum, recall KE of photo electron (not of the metal) = energy of photon falling - work function
That is KE = hv - W
Or KE = hc/L - hvo
Printing inconvenience because of lack of waned symbols
let us compute h c / L = 6.626 x 10^-34 * 3 x 10^8 / 2000 x 10^-10 = 9.939 x 10^-19 J
To convert J into eV let us divide by e = 1.6 x 10^-19 J
So we get energy of photon falling = 6.212 e V
Therefore KE of photo electron = 6.212 - 4.2 = 2 eV (because of only 2 significant figure in 4.2 I have reduced for the other)
If needed in J simply multiply by e = 1.6 x 10^-19 J
That is KE = hv - W
Or KE = hc/L - hvo
Printing inconvenience because of lack of waned symbols
let us compute h c / L = 6.626 x 10^-34 * 3 x 10^8 / 2000 x 10^-10 = 9.939 x 10^-19 J
To convert J into eV let us divide by e = 1.6 x 10^-19 J
So we get energy of photon falling = 6.212 e V
Therefore KE of photo electron = 6.212 - 4.2 = 2 eV (because of only 2 significant figure in 4.2 I have reduced for the other)
If needed in J simply multiply by e = 1.6 x 10^-19 J
