AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The expression for work function for Li is, $v_{0}= \frac{W_{0}}{h}$ Here, the threshold frequency is given as $v_{0}$ and the work function is $W_{0}$ Substitute the given value in above expression. $v_{0}= \frac{2.42 \times 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ $= 0.584 \times 10^{15} \mathrm{s}^{-1}$ The expression for work function for K is, $v_{0}= \frac{W_{0}}{h}$ Substitute the given value in above expression. $v_{0}= \frac{2.25 \times 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ $= 0.54 \times 10^{15} \mathrm{s}^{-1}$ The expression for work function for Mg is, $v_{0}= \frac{W_{0}}{h}$ Substitute the given value in above expression. $v_{0}= \frac{3.70 \times 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ $= 0.89 \times 10^{15} \mathrm{s}^{-1}$ The expression for work function for Ag is, $v_{0}= \frac{W_{0}}{h}$ Substitute the given value in above expression. $v_{0}= \frac{4.30 \times 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ $= 1.03 \times 10^{15} \mathrm{s}^{-1}$ The expression for work function for Cu is, $v_{0}= \frac{W_{0}}{h}$ Substitute the given value in above expression. $v_{0}= \frac{4.80 \times 1.602 \times 10^{-19} \mathrm{J}}{6.626 \times 10^{-34} \mathrm{Js}}$ $= 1.16 \times 10^{15} \mathrm{s}^{-1}$ The frequency of light is given as, $v= \frac{c}{\lambda}= \frac{3 \times 10^{8}}{540 \times 10^{-9}}$ $= 0.56 \times 10^{15} \mathrm{s}^{-1}$ The condition at which the photoelectric effect is shown is $v>v_{0},$ which is only observed in case of Potassium.