Saved Bookmarks
| 1. |
When light of wavelength 2200 Å falls on Cu, photo electrons are emitted from it. Find (i) the threshold wavelength and (ii) the stopping potential. Given : the work function for Cu is phi_(0) = 4.7 eV. |
|
Answer» Solution :(i) The threshold wavelength is given by `lambda_(0) = (hc)/(phi_(0)) = (6.626 xx 10^(-43) xx 3 xx 10^(8))/(4.7 xx 1.6 xx 10^(-19)) = 2643 Å` (II) ENERGY of the photon of wavelength 2200 `Å` is `E = (hc)/(lambda) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(2200 xx 10^(-10))` `= 9.035 xx 10^(-19)` J = 5.65 eV We know that kinetic energy of fastest photo ELECTRON is `K_(MAX) = h upsilon - phi_(0) = 5.65 - 4.7 = 0.95 eV` From equation, `K_(max) = eV_(0)` `V_(0) = (K_(max))/(e) = (0.95 xx 1.6 xx 10^(-19))/(1.6 xx 10^(-19))` Therefore, stopping potential = 0.95 V |
|