Verified Solution • 2025–26 Session moderate

QPages 365-366: Q3: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q3 Solution: An atom absorbs a photon of wavelength 500 nm and emits another …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently reverse the $E=hc/\lambda$ relationship, assuming a longer wavelength carries more energy or adding the two photon energies instead of subtracting the emitted one. Always compute each photon’s energy as $E=1240/\lambda\;(\text{eV})$ (with $\lambda$ in nm) and take net absorbed = $E_{\text{absorbed}}-E_{\text{emitted}}$.

An atom absorbs a photon of wavelength 500 nm and emits another photon of wavelength 700 nm. Find the net energy absorbed by the atom in the process. 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:** Net energy absorbed = 0.71 eV. **Step-by-Step Explanation:** Energy absorbed: E1=1240500=2.48E_1 = \frac{1240}{500} = 2.48 eV. Energy emitted: E2=1240700=1.77E_2 = \frac{1240}{700} = 1.77 eV. Net energy absorbed =2.48−1.77=0.71= 2.48 - 1.77 = 0.71 eV.

All Questions in this Chapter

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