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 $(\lambda)$ of incident light and the corresponding stopping potential $(V_{0})$ are given below: | $\lambda(\mu \mathrm{m})$ | $V_{0}$ (Volt) | | --- | --- | | 0.3 | 2.0 | | 0.4 | 1.0 | | 0.5 | 0.4 | Given that $c=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$ and $e=1.6 \times 10^{-19} \mathrm{C}$, Planck's constant (in units of J s) found from such an experiment is

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Modern Physics

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:

$λ (μ m)$	$V_{0}$ (Volt)
0.3	2.0
0.4	1.0
0.5	0.4

Given that

c = 3 \times 1 0^{8} m s^{- 1}

and

e = 1.6 \times 1 0^{- 19} C

, Planck's constant (in units of J s) found from such an experiment is

(A) $6.0 \times 1 0^{- 34}$
(B) $6.4 \times 1 0^{- 34}$ ✓ Correct
(C) $6.6 \times 1 0^{- 34}$
(D) $6.8 \times 1 0^{- 34}$

Answer: (B)

Solution

From the photoelectric equation,

e V = \frac{h c}{λ} - ϕ

. For two wavelengths:

e (V_{1} - V_{2}) = \frac{h c}{λ _{1}} - \frac{h c}{λ _{2}} = \frac{h c ( λ _{2} - λ _{1} )}{λ _{1} λ _{2}}

Solving for

h

h = \frac{e ( V _{1} - V _{2} ) λ _{1} λ _{2}}{c ( λ _{2} - λ _{1} )}

Using the first two data points (

λ_{1} = 0.3 μ m

V_{1} = 2.0

λ_{2} = 0.4 μ m

V_{2} = 1.0

V):

h = \frac{1.6 \times 1 0 ^{- 19} \times 1.0 \times 0.3 \times 0.4 \times 1 0 ^{- 12}}{3 \times 1 0 ^{8} \times 0.1 \times 1 0 ^{- 6}} = 6.4 \times 1 0^{- 34}

J s

Alternatively, the slope of

V_{0}

1/ λ

equals

h c / e

, which gives the same result.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option

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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Modern Physics

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