21mv12=hf1−ϕ
21mv22=hf2−ϕ
v12−v22=m2h(f1−f2)
JEE Main 2021 — Physics Modern Physics
Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-electrons coming out are v1 and v2 respectively, then
Held on 17 Mar 2021 · Verified 6 Jul 2026.
v12−v22=m2h[f1−f2]
v12+v22=m2h[f1+f2]
v1+v2=[m2h(f1+f2)]21
v1−v2=[m2h(f1−f2)]21
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