A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?
Solution
$k{E_{\max }} = {{hc} \over {{\lambda _i}}} + \phi$<br><br>or $e{V_o} = {{hc} \over {{\lambda _i}}} + \phi$<br><br>when $\lambda$<sub>i</sub> = 670.5 nm ; V<sub>o</sub> = 0.48<br><br>when $\lambda$<sub>i</sub> = 474.6 nm ; V<sub>o</sub> = ?<br><br>So, <br><br>$e(0.48) = {{1240} \over {670.5}} + \phi$ ..... (1)<br><br>$e({V_o}) = {{1240} \over {474.6}} + \phi$ .....(2)<br><br>(2) $-$ (1)<br><br>$e({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV$<br><br>${V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right)$ Volts<br><br>V<sub>o</sub> = 0.48 + 0.76<br><br>V<sub>o</sub> = 1.24 V $\simeq$ 1.25 V
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%.