JEE Challenger
More from Dual Nature of Radiation and Matter
PhyMedAdv 2016 • Paper 1

Determination of Plancks Constant from Photoelectric Effect Data

In a historical experiment performed 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) and the corresponding stopping potential (V0V_0) are given below:

λ (\mum)V0 (Volt)0.32.00.41.00.50.4\begin{array}{cc} \lambda \text{ (\mu m)} & V_0 \text{ (Volt)} \\ \hline 0.3 & 2.0 \\ 0.4 & 1.0 \\ 0.5 & 0.4 \end{array}

Given that c=3×108 m s−1c = 3 \times 10^8 \text{ m s}^{-1} and e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}, Planck's constant (in units of J s\text{J s}) found from such an experiment is

Options

A

6.0×10−346.0 \times 10^{-34}

B

6.4×10−346.4 \times 10^{-34}

Correct
C

6.6×10−346.6 \times 10^{-34}

D

6.8×10−346.8 \times 10^{-34}

Topics & Concepts

Dual Nature of Radiation and MatterPhotoelectric Effect

Step-by-Step Solution

According to Einstein's photoelectric equation, the relationship between the stopping potential V0V_0, the wavelength of incident light λ\lambda, the work function of the metal Φ\Phi, and Planck's constant hh is given by:

eV0=hcλ−Φe V_0 = \frac{hc}{\lambda} - \Phi

To eliminate the work function Φ\Phi, we can take two different data points from the given table:

Data Point 1: λ1=0.3 μm=0.3×10−6 m,V01=2.0 V\lambda_1 = 0.3 \, \mu\text{m} = 0.3 \times 10^{-6} \text{ m}, \quad V_{01} = 2.0 \text{ V}

Data Point 2: λ2=0.4 μm=0.4×10−6 m,V02=1.0 V\lambda_2 = 0.4 \, \mu\text{m} = 0.4 \times 10^{-6} \text{ m}, \quad V_{02} = 1.0 \text{ V}

Subtracting the equation for Data Point 2 from Data Point 1:

e(V01−V02)=hc(1λ1−1λ2)e(V_{01} - V_{02}) = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)

Substituting the given values (c=3×108 m s−1c = 3 \times 10^8 \text{ m s}^{-1} and e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}):

1.6×10−19×(2.0−1.0)=h×(3×108)×(10.3×10−6−10.4×10−6)1.6 \times 10^{-19} \times (2.0 - 1.0) = h \times (3 \times 10^8) \times \left( \frac{1}{0.3 \times 10^{-6}} - \frac{1}{0.4 \times 10^{-6}} \right)

1.6×10−19=h×3×1014×(103−104)1.6 \times 10^{-19} = h \times 3 \times 10^{14} \times \left( \frac{10}{3} - \frac{10}{4} \right)

1.6×10−19=h×3×1014×(56)1.6 \times 10^{-19} = h \times 3 \times 10^{14} \times \left( \frac{5}{6} \right)

1.6×10−19=2.5×1014×h1.6 \times 10^{-19} = 2.5 \times 10^{14} \times h

Solving for hh:

h=1.6×10−192.5×1014=0.64×10−33 J s=6.4×10−34 J sh = \frac{1.6 \times 10^{-19}}{2.5 \times 10^{14}} = 0.64 \times 10^{-33} \text{ J s} = 6.4 \times 10^{-34} \text{ J s}

We can also check using Data Point 2 and Data Point 3 (λ3=0.5×10−6 m,V03=0.4 V\lambda_3 = 0.5 \times 10^{-6} \text{ m}, V_{03} = 0.4 \text{ V}):

e(V02−V03)=hc(1λ2−1λ3)e(V_{02} - V_{03}) = hc \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_3} \right)

1.6×10−19×(1.0−0.4)=h×(3×108)×(10.4×10−6−10.5×10−6)1.6 \times 10^{-19} \times (1.0 - 0.4) = h \times (3 \times 10^8) \times \left( \frac{1}{0.4 \times 10^{-6}} - \frac{1}{0.5 \times 10^{-6}} \right)

0.96×10−19=h×3×1014×(2.5−2.0)0.96 \times 10^{-19} = h \times 3 \times 10^{14} \times (2.5 - 2.0)

0.96×10−19=1.5×1014×h0.96 \times 10^{-19} = 1.5 \times 10^{14} \times h

h=0.96×10−191.5×1014=6.4×10−34 J sh = \frac{0.96 \times 10^{-19}}{1.5 \times 10^{14}} = 6.4 \times 10^{-34} \text{ J s}

Thus, the value of Planck's constant determined from the experimental data is 6.4×10−34 J s6.4 \times 10^{-34} \text{ J s}.

Correct Option: (B)

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