When radiation of wavelength $\lambda$ is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is :
Solution
${V_s} = hv - \phi$<br><br>$4.8 = {{hc} \over \lambda } - \phi$ ..... (i)<br><br>$1.6 = {{hc} \over {2\lambda }} - \phi$ ..... (ii)<br><br>Using above equation (i) - (ii)<br><br>$3.2 = {{hc} \over \lambda } - {{hc} \over {2\lambda }}$<br><br>$3.2 = {{hc} \over {2\lambda }}$ ..... (iii)<br><br>$\left[ {\lambda = {{hc} \over {6.4}}} \right]$<br><br>Put in equation (ii)<br><br>$\phi$ = 1.6<br><br>${{hc} \over {{\lambda _{th}}}} = 1.6$<br><br>${\lambda _{th}} = {{hc} \over {1.6}}$<br><br>$= \left( {{{hc} \over {6.4}}} \right) \times 4 = 4\lambda$
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
This question is part of PrepWiser's free JEE Main question bank. 145 more solved questions on Dual Nature of Matter and Radiation are available — start with the harder ones if your accuracy is >70%.