Verified Solution • 2025–26 Session moderate

QJEEAdv16Phbae1: JEE Advanced 2016 Physics - Class 12 JEEAdv16Phbae1 Solution: In a historical experiment to determine Planck's constant, a…

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\lambda ) of incident light and the corresponding stopping potential (V0) are given below:

λ(μm)\lambda \left( {\mu m} \right) V0(Volt)
0.3 2.0
0.4 1.0
0.5 0.4


Given that c = 3 ×\times 108 ms-1 and e = 1.6 ×\times 10-19 C, Planck's constant (in units of J-s) found from such an experiment is) :

Multiple Choice Options

  • A 6.0 ×\times 10-34
  • B 6.6 ×\times 10-34
  • C 6.4 ×\times 10-34
  • D 6.8 ×\times 10-34

Step-by-Step Verified Solution

hcλ−ϕ=eV0{{hc} \over \lambda } - \phi = e{V_0} ( ϕ\phi = work function)

hc0.3×10−6−ϕ=2e{{hc} \over {0.3 \times {{10}^{ - 6}}}} - \phi = 2e .... (i)

hc0.4×10−6−ϕ=1e{{hc} \over {0.4 \times {{10}^{ - 6}}}} - \phi = 1e .... (ii)

Subtracting Eq. (ii) from Eq. (i)

hc(10.3−10.4)106=ehc\left( {{1 \over {0.3}} - {1 \over {0.4}}} \right){10^6} = e

hc(0.10.12×106)=ehc\left( {{{0.1} \over {0.12}} \times {{10}^6}} \right) = e

h=0.64×10−33h = 0.64 \times {10^{ - 33}}

=6.4×10−34= 6.4 \times {10^{ - 34}} J-s

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