JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 11)
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)
1.3 V
1.1 V
1.9 V
0.6 V
Explanation
$$K{E_{\max }} = e{V_s} = {{hc} \over \lambda } - \phi $$
$$ \Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 = $$ 1.93 eV
$$ \Rightarrow {V_{{s_1}}} = $$ 1.93 V .... (i)
$$ \Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 = $$ 0.6 eV
$$ \Rightarrow {V_{{s_2}}} = $$ 0.6 V .... (ii)
$$\Delta$$V = $${V_{{s_1}}} - {V_{{s_2}}}$$ = 1.93 $$-$$ 0.6 = 1.33 V
$$ \Rightarrow e{V_s} = {{1240} \over {280}} - 2.5 = $$ 1.93 eV
$$ \Rightarrow {V_{{s_1}}} = $$ 1.93 V .... (i)
$$ \Rightarrow e{V_{{s_2}}} = {{1240} \over {400}} - 2.5 = $$ 0.6 eV
$$ \Rightarrow {V_{{s_2}}} = $$ 0.6 V .... (ii)
$$\Delta$$V = $${V_{{s_1}}} - {V_{{s_2}}}$$ = 1.93 $$-$$ 0.6 = 1.33 V
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