Question 7 of 10intermediate🔧 ApplyNumerical3 marks

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×4.12 \times 10^{-15}$$ Vs. Calculate the value of Planck's constant.

Correct Answer

Planck's constant h = 6.59×106.59 \times 10⁻³⁴ J s

Exercise: EXERCISES | Q: 11.5 | (Chapter: 285)
For More Understanding

Explanation

This question tests the understanding of Einstein's photoelectric equation and the relationship between stopping potential and frequency. From the equation V₀e = hν - φ₀, when we plot cut-off voltage (V₀) versus frequency (ν), we get a straight line. The slope of this line equals h/e, where h is Planck's constant and e is the electronic charge. Millikan used this relationship to determine the value of Planck's constant experimentally.

Solution Steps

  1. Step 1: Identify the relationship between slope and Planck's constant. From Einstein's photoelectric equation: V₀e = hν - φ₀ Rearranging: V₀ = (h/e)ν - φ₀/e This is of the form y = mx + c, where slope m = h/e Therefore, slope = h/e

  2. Step 2: Substitute the given values. Given: Slope = 4.12×104.12 \times 10⁻¹⁵ Vs Electronic charge, e = 1.6×101.6 \times 10⁻¹⁹ C From slope = h/e: h = slope × e

  3. Step 3: Calculate Planck's constant. h = (4.12×104.12 \times 10⁻¹⁵ Vs) × (1.6×101.6 \times 10⁻¹⁹ C) h = 6.592×106.592 \times 10⁻³⁴ J s h ≈\approx 6.59×106.59 \times 10⁻³⁴ J s

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