Verified Solution • 2025–26 Session moderate

QPages 395-396: Q26: HC Verma: Concepts of Physics - Volume 2 X-Rays Pages 395-396: Q26 Solution: The stopping potential in a photoelectric experiment is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often replace the frequency ν with 1/λ directly, forgetting that ν = c/λ; this omits the factor of the speed of light and yields a slope of $h/e$ instead of the correct $hc/e$. Always include the $c$ when converting between frequency and inverse wavelength to obtain the proper slope.

The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength 1/λ1/\lambda) of the light falling on the cathode. The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength 1/λ1/\lambda) of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find its value. Use Planck constant h=6.63×10−34 Js=4.14×10−15 eVsh = 6.63 \times 10^{-34} \text{ Js} = 4.14 \times 10^{-15} \text{ eVs}, speed of light c=3×108 m/sc = 3 \times 10^8 \text{ m/s}.)

Step-by-Step Verified Solution

**Answer:** Slope = h/e=4.14×10−15h/e = 4.14 \times 10^{-15} V-s. **Step-by-Step Explanation:** For photoelectric effect: eVs=hν−ϕeV_s = h\nu - \phi. So Vs=heν−ϕeV_s = \frac{h}{e}\nu - \frac{\phi}{e}. For X-ray tube: eV=hνmax=hcλmineV = h\nu_{max} = \frac{hc}{\lambda_{min}}. So V=hce⋅1λminV = \frac{hc}{e} \cdot \frac{1}{\lambda_{min}}. Plotting VV vs 1/λ1/\lambda gives slope hc/ehc/e. Plotting VsV_s vs ν\nu gives slope h/eh/e. The slopes are related: (h/e)×c=hc/e(h/e) \times c = hc/e. Both slopes equal h/eh/e in appropriate units. Slope = h/e=6.63×10−341.6×10−19=4.14×10−15h/e = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} = 4.14 \times 10^{-15} V-s.

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