Verified Solution • 2025–26 Session moderate

QPages 365-366: Q14: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q14 Solution: Find the maximum kinetic energy of the photoelectrons ejected…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently insert the wavelength in metres into the shortcut $E( ext{eV})=1240/\lambda(\text{nm})$, producing an answer off by $10^9$. Always convert the wavelength to nanometres before using the 1240 constant, or use $E=hc/\lambda$ with consistent SI units, then subtract the work function in the same energy units.

Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350 nm is incident on a cesium surface. Work function of cesium = 1.9 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=1.64KE_{max} = 1.64 eV. **Step-by-Step Explanation:** Energy of photon: E=1240350=3.54E = \frac{1240}{350} = 3.54 eV. KEmax=E−ϕ=3.54−1.9=1.64KE_{max} = E - \phi = 3.54 - 1.9 = 1.64 eV.

All Questions in this Chapter

Related Learning Resources & Exam Preparation