Let ϕ= work function of the metal, λ1hc=ϕ+eV1 λ2hc=ϕ+eV2…… (ii) Sutracting (ii) from (i) we get hc(λ11−λ21)=e(V1−V2) ⇒V1−V2=ehc(λ1⋅λ2λ2−λ1)λ1=300nmλ2=400nmehc=1240nm−V =(1240nm−v)(300nm×400nm100nm) =1,03 V≈1 V
JEE Main 2019 — Physics Modern Physics
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300nm to 400nm. The decrease in the stopping potential is close to : (ehc=1240nm−V)
Held on 11 Jan 2019 · Verified 6 Jul 2026.
0.5 V
1.5 V
1.0 V
2.0 V
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