Kinetic energy KE=ϕ−ϕ0 Here, KE1=1−0.5=0.5eV KE2=2.5−0.5=2eV∴KE2KE1=20.5=41 or y22y12=41 or y2y1=41=21
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic surface whose work function is 0.5…
NEET UG 2011 — Physics Modern Physics
2011mcqmedium
Light of two different frequencies whose photons have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5eV successively. Ratio of maximum speeds of emitted electrons will be
Official previous-year question
Held on 30 Apr 2011 · Verified 9 Jul 2026.
Options
- A
1:2
- B
1:1
- C
1:5
- D
1:4
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