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QPages 365-366: Q17: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q17 Solution: Find the maximum magnitude of the linear momentum of a…

Find the maximum magnitude of the linear momentum of a photoelectron emitted when a wavelength of 400 nm falls on a metal with work function 2.5 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:** pmax=6.6×10−25 kg m/sp_{max} = 6.6 \times 10^{-25} \text{ kg m/s}. **Step-by-Step Explanation:** Photon energy: E=1240400=3.1E = \frac{1240}{400} = 3.1 eV. KEmax=3.1−2.5=0.6KE_{max} = 3.1 - 2.5 = 0.6 eV =9.6×10−20= 9.6 \times 10^{-20} J. pmax=2mKE=2×9.1×10−31×9.6×10−20=4.18×10−25p_{max} = \sqrt{2mKE} = \sqrt{2 \times 9.1 \times 10^{-31} \times 9.6 \times 10^{-20}} = 4.18 \times 10^{-25} kg m/s.

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