Verified Solution • 2025–26 Session moderate

QPages 365-366: Q2: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q2 Solution: Calculate the momentum of a photon of light of wavelength 500…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the wavelength from nanometers to meters before applying $p = h/\lambda$, leading to an answer that is off by a factor of $10^9$. Always ensure $\lambda$ is in SI units (meters) when using $h$ in J-s to obtain momentum in kg m/s.

Calculate the momentum of a photon of light of wavelength 500 nm. 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:** p=1.33×10−27 kg m/sp = 1.33 \times 10^{-27} \text{ kg m/s}. **Step-by-Step Explanation:** p=hλ=6.63×10−34500×10−9=1.33×10−27 kg m/sp = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{500 \times 10^{-9}} = 1.33 \times 10^{-27} \text{ kg m/s}.

All Questions in this Chapter

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