question_answer 70)

Work function of caesium metal is 1.9 eV. Calculate: (a) the threshold wavelength (b) the threshold frequency of the radiation (c) If the caesium metal is irradiated with a wavelength of 500 nm, calculate the kinetic energy and velocity of the ejected electron.

Answer:

(a) \[{{\lambda }_{0}}=\frac{c}{{{v}_{0}}}=\frac{3\times {{10}^{8}}}{4.587\times {{10}^{14}}}\] \[=654\times {{10}^{-9}}m\] \[=654nm\] (b) \[{{v}_{0}}=\frac{{{w}_{0}}}{h}\frac{1.9\times 1.6\times {{10}^{-19}}}{6.626\times {{10}^{-34}}}\] \[=4.587\times {{10}^{14}}{{s}^{-1}}\] (c) Kinetic energy of photoelectrons = Absorbed energy - Threshold energy \[=\frac{hc}{\lambda }-{{w}_{0}}\] \[=\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{500\times {{10}^{-9}}}-1.9\times 1.6\times {{10}^{-19}}\] \[=9.36\times {{10}^{-20}}J\] \[\frac{1}{2}m{{\upsilon }^{2}}=9.36\times {{10}^{-20}}\] \[\frac{1}{2}\times 9.1\times {{10}^{-31}}\times {{\upsilon }^{2}}=9.36\times {{10}^{-20}}\] \[\upsilon =4.53\times {{10}^{5}}m{{s}^{-1}}\]