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QJEEM20Ph2792: JEE Main 2020 Physics - Class 12 JEEM20Ph2792 Solution: The surface of a metal is illuminated alternately with photons…

The surface of a metal is illuminated alternately with photons of energies E1 = 4 eV and E2 = 2.5 eV respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is _____.

Step-by-Step Verified Solution

Given,

E1 = 4 eV

E2 = 2.5 eV

and (V1)max⁡(V2)max⁡=2{{{{\left( {{V_1}} \right)}_{\max }}} \over {{{\left( {{V_2}} \right)}_{\max }}}} = 2

12m((V1)max⁡)212m((V2)max⁡)2=E1−ϕ0E2−ϕ0{{{1 \over 2}m{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{1 \over 2}m{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{{E_1} - {\phi _0}} \over {{E_2} - {\phi _0}}}

⇒\Rightarrow ((V1)max⁡)2((V2)max⁡)2=4−ϕ02.5−ϕ0{{{{\left( {{{\left( {{V_1}} \right)}_{\max }}} \right)}^2}} \over {{{\left( {{{\left( {{V_2}} \right)}_{\max }}} \right)}^2}}} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}

⇒\Rightarrow (2)2=4−ϕ02.5−ϕ0{\left( 2 \right)^2} = {{4 - {\phi _0}} \over {2.5 - {\phi _0}}}

⇒\Rightarrow 10 - 4 ϕ\phi 0 = 4 - ϕ\phi 0 ⇒\Rightarrow 3 ϕ\phi 0 = 6

⇒\Rightarrow ϕ\phi 0 = 2

∴\therefore Work function( ϕ\phi ) of the metal = 2 eV

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