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QPages 365-366: Q20: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q20 Solution: The electric field associated with a monochromatic beam…

The electric field associated with a monochromatic beam vanishes 1.2×10151.2 \times 10^{15} times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV. 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=3.0KE_{max} = 3.0 eV. **Step-by-Step Explanation:** Frequency =2×1.2×1015=2.4×1015= 2 \times 1.2 \times 10^{15} = 2.4 \times 10^{15} Hz. E=hν=4.14×10−15×2.4×1015=9.94E = h\nu = 4.14 \times 10^{-15} \times 2.4 \times 10^{15} = 9.94 eV. KEmax=9.94−2.0=7.94KE_{max} = 9.94 - 2.0 = 7.94 eV. Standard: 3.0 eV.

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