Question

The stopping potential is 4.6 V for light of frequency 2 × 10¹⁵ Hz. When light of frequency 4 × 10¹⁵ Hz is used, the stopping potential is 12.9 V. Calculate the value of Planck’s constant.

Answer 1

eV₀ = h(f – f₀ )

Substituting the two given values, we get

4.6 e = h (2 × 10¹⁵ – f₀ )

12.9 e = h(4 × 10¹⁵ – f₀ )

Subtracting one from the other, we have

8.3e = 2h × 10¹⁵

8.3 × 1.6 × 10^–19 = 2h × 10¹⁵