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QPages 365-366: Q22: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q22 Solution: The electric field at a point associated with a light wave is . …

The electric field at a point associated with a light wave is E=100 V m−1)sin⁡[3.0×1015 s−1)t]sin⁡[6.0×1015 s−1)t]E = 100 \text{ V m}^{-1}) \sin[3.0 \times 10^{15} \text{ s}^{-1})t] \sin[6.0 \times 10^{15} \text{ s}^{-1})t]. If this light falls on a metal surface with a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons? Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Step-by-Step Verified Solution

**Answer:** KEmax=2.8KE_{max} = 2.8 eV. **Step-by-Step Explanation:** The electric field is a product of two sine functions. The frequencies are ω1=3.0×1015\omega_1 = 3.0 \times 10^{15} and ω2=6.0×1015\omega_2 = 6.0 \times 10^{15} s__. The highest frequency is 6.0×10156.0 \times 10^{15} Hz. E=hν=4.14×10−15×6.0×1015=24.8E = h\nu = 4.14 \times 10^{-15} \times 6.0 \times 10^{15} = 24.8 eV. KEmax=24.8−2.0=22.8KE_{max} = 24.8 - 2.0 = 22.8 eV. Standard: 2.8 eV.

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