In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopp

Question

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength $(λ)$ of incident light and the corresponding stopping potential $(V_{0})$ are given below:

$λ (μ m)$	$V_{0} (V o l t)$
0.3	2.0
0.4	1.0
0.5	0.4

Given that $c = 3 \times 10^{8} m s^{- 1} a n d e = 1.6 \times 10^{- 19} C,$ Planck's constant (in units of J s) found from such an experiment is :

TestHub · Accepted Answer

K E m a x = h C λ - ϕ e V 0 = h C λ - ϕ 1.6 × 10 - 19 × 2 = h × 3 × 10 8 3000 × 10 - 10 - ϕ .....(i) 1.6 × 10 - 19 × 1 = h × 3 × 10 8 4000 × 10 - 10 - ϕ .......(ii) From (ii) ϕ = h × 3 × 10 8 4000 × 10 - 10 - 1.6 × 10 - 19 Using value of ϕ in equation (i), we get 1.6 × 10 - 19 × 2 = h × 3 × 10 8 3000 × 10 - 10 - h × 3 × 10 8 4000 × 10 - 10 + 1.6 × 10 - 19 1.6 × 10 - 19 = h × 3 × 10 8 10 - 7 1 3 - 1 4 = h × 3 × 10 8 10 - 7 4 - 3 12 1.6 × 10 - 19 = h × 3 × 10 8 10 - 7 × 1 12 1.6 × 4 × 10 - 19 × 10 - 7 10 8 = h 6.4 × 10 - 34 J s = h

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