Question

In the arrangement shown in the figure, y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. Find the stopping potential V needed to stop the photocurrent.
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Answer 1

Explanation:

To find: The stopping potential V needed to stop the photocurrent.

Given data in the question,  

Width of Fringe, y = 1 mm ×2 = 2 mm

Function of Work, $W_{0}=2.2 \mathrm{eV}$

D = 1.2 m

d = 0.24 mm

Width of Fringe,

$y=\frac{\lambda D}{d}$ , λ = light wavelength

$\lambda=\frac{2 \times 10^{-3} \times 0.24 \times 10^{-3}}{1.2}$

  $=4 \times 10^{-7} \mathrm{m}$

Energy,

$E=\frac{h c}{\lambda}$

    $=\frac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{4 \times 10^{-7}}$

    = 3.105 eV

From the Photoelectric Equation of Einstein,

$e V_{0}=E-W_{0}$

Where $V_O$ is the halt potential and where e is electron charge.

$e V_{0}=3.105-2.2$

      = 0.905 eV

 $V_{0}=\frac{0.905}{1.6 \times 10^{-19}} \times 1.6 \times 10^{-19} \mathrm{V}$

      = 0.905 V