Dual Nature of Matter

JEE Main level, single correct, medium

145 of 200

Question 145 of 200

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−1E=200\left[\sin \left(6\times {10}^{15}\right)t+\sin \left(9\times {10}^{15}\right)t\right]{Vm}^{-1}
Given: h=4.14×10−15eVsh=4.14\times {10}^{-15}eVs
If this light falls on a metal surface having a work function of 2.50eV2.50eV, the maximum kinetic energy of the photoelectrons will be

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Answer: (D) 3.42eV3.42eV

Given: E=200[sin⁡(6×1015)t+sin⁡(9×1015)t]Vm−1E=200\left[\sin \left(6\times {10}^{15}\right)t+\sin \left(9\times {10}^{15}\right)t\right]V{m}^{-1}
Here, angular velocity, ω1=6×1015{\omega }_{1}=6\times {10}^{15}and ω2=9×1015{\omega }_{2}=9\times {10}^{15}
Maximum kinetic energy of the photoelectron is KEmax⁡.=E−ϕ{KE}_{\max .}=E-\phi
=hf−ϕ=hω2π−ϕ=hf-\phi =\dfrac{h\omega }{2\pi }-\phi
=4.14×10−15×9×10152×3.14−2.5=5.92−2.50=3.42eV=\dfrac{4.14\times {10}^{-15}\times 9\times {10}^{15}}{2\times 3.14}-2.5=5.92-2.50=3.42eV

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