Given: E=200[sin(6×1015)t+sin(9×1015)t]Vm−1
Here, angular velocity, ω1=6×1015and ω2=9×1015
Maximum kinetic energy of the photoelectron is KEmax.=E−ϕ
=hf−ϕ=2πhω−ϕ
=2×3.144.14×10−15×9×1015−2.5=5.92−2.50=3.42eV
The electric field at a point associated with a light wave is given by E=200[ sin(6× 10^15)t+ sin(9× 10^15)t] Vm^-1 Given: h=4.14× 10^-15 eVs If this…
JEE Main 2022 — Physics Modern Physics
2022mcqeasy
The electric field at a point associated with a light wave is given by
E=200[sin(6×1015)t+sin(9×1015)t]Vm−1
Given: h=4.14×10−15eVs
If this light falls on a metal surface having a work function of 2.50eV, the maximum kinetic energy of the photoelectrons will be
Official previous-year question
Held on 29 Jun 2022 · Verified 6 Jul 2026.
Options
- A
1.90eV
- B
3.27eV
- C
3.60eV
- D
3.42eV
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