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QPages 365-366: Q19: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q19 Solution: In an experiment on photoelectric effect, the stopping…

⚠️ Common Exam Traps & Conceptual Pitfalls

When plotting $V$ versus $1/lambda$, the independent variable must be expressed in ext{m}^{-1}. If $1/lambda$ is taken in ext{nm}^{-1}$, the slope will come out in V·nm, which cannot be directly used in $h= rac{ ext{slope},e}{c}$ because $c$ is in ext{m/s}. Convert $1/lambda$ to ext{m}^{-1}$ (or multiply the slope by $10^{-9}$) before applying the formula. Also remember that the intercept equals $-phi/e$, so the work function is $- ext{intercept} imes e$, not the intercept itself.

In an experiment on photoelectric effect, the stopping potential is measured for monochromatic light beams corresponding to different wavelengths. The data collected are as follows:- Wavelength nm): 350, 400, 450, 500, 550; Stopping potential V): 1.45, 1.00, 0.66, 0.38, 0.16. Plot the stopping potential against inverse of wavelength 1/λ1/\lambda) on a graph paper and find a) Planck's constant b) the work function of the emitter and c) the threshold wavelength. c) the threshold wavelength. 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:** (a) h=6.6×10−34h = 6.6 \times 10^{-34} J-s (b) ϕ=2.0\phi = 2.0 eV (c) λ0=620\lambda_0 = 620 nm. **Step-by-Step Explanation:** Plot VsV_s vs 1/λ1/\lambda. Slope =hc/e= hc/e. From data: slope ≈1.24×10−6\approx 1.24 \times 10^{-6} V�m. h=slope×ec=1.24×10−6×1.6×10−193×108=6.6×10−34h = \frac{\text{slope} \times e}{c} = \frac{1.24 \times 10^{-6} \times 1.6 \times 10^{-19}}{3 \times 10^8} = 6.6 \times 10^{-34} J-s. Intercept gives ϕ=2.0\phi = 2.0 eV, λ0=620\lambda_0 = 620 nm.

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