In a photoelectric experiment ultraviolet light of wavelength 280 nm is used with lithium cathode having work function $\phi$ = 2.5 eV. If the wavelength of incident light is switched to 400 nm, find out the change in the stopping potential. (h = 6.63 $\times$ 10$-$34 Js, c = 3 $\times$ 108 ms$-$1)
Solution
$K{E_{\max }} = e{V_s} = {{hc} \over \lambda } - \phi$<br><br>$\Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 =$ 1.93 eV<br><br>$\Rightarrow {V_{{s_1}}} =$ 1.93 V .... (i)<br><br>$\Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 =$ 0.6 eV<br><br>$\Rightarrow {V_{{s_2}}} =$ 0.6 V .... (ii)<br><br>$\Delta$V = ${V_{{s_1}}} - {V_{{s_2}}}$ = 1.93 $-$ 0.6 = 1.33 V
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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