question_answer 138)

  Threshold frequency, \[{{v}_{0}}\] is the minimum frequency which a photon must possess to eject an electron form a metal. It is different for different metals. When a photon of frequency \[1.0\times {{10}^{15}}{{s}^{-1}}\] was allowed to hit a metal surface, an electron having \[1.988\times {{10}^{-19}}J\] of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

Answer:

  Absorbed Energy = Threshold + Kinetic Energy Energy   of photoelectron \[hv=h{{v}_{0}}+K.E.\]             \[6.626\times {{10}^{-34}}\times 1\times {{10}^{15}}=6.626\times {{10}^{-34}}\times {{v}_{0}}+1.988\times {{10}^{-19}}J\]                 \[{{v}_{0}}=\frac{(6.626\times {{10}^{19}}-1.988\times {{10}^{-19}})}{6.626\times {{10}^{-34}}}\]                 \[=6.99\times {{10}^{14}}Hz\] Frequency of photon striking metal surface: \[v=\frac{c}{\lambda }=\frac{3\times {{10}^{8}}}{600\times {{10}^{-9}}}\] \[=5\times {{10}^{14}}Hz\] Since \[v<{{v}_{0}}\] hence photoeleetron will not be ejected.