Ncert Exercises & Solutions

Question 11.27: Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon lamp irradiates photosensitive material made of cesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

Hint: Use Einstein's photoelectric equation.
Step 1: Find the work function.

$\begin{aligned} ϕ_{0} & = \frac{hc}{λ} - {eV}_{0} \\ = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{640 .2 \times 10^{- 9}} - 1 .6 \times 10^{- 19} \times 0 .54 \\ = 3 .093 \times 10^{- 19} - 0 .864 \times 10^{- 19} \\ = 2 .229 \times 10^{- 19} J \\ = \frac{2 .229 \times 10^{- 19}}{1 .6 \times 10^{- 19}} = 1 .39 eV \end{aligned}$

Step 2: Find the new stopping potential.
The wavelength of the radiation emitted from an iron source, $λ^{'} = 427 .2 nm = 427 .2 \times 10^{- 9} m$

Let V₀ be the new stopping potential. Hence, photo-energy is given as:

$\begin{aligned} {eV}_{0}^{'} & = \frac{hc}{λ^{'}} - ϕ_{0} \\ = \frac{6 .6 \times 10^{- 34} \times 3 \times 10^{8}}{427 .2 \times 10^{- 9}} - 2 .229 \times 10^{- 19} \\ = 4 .63 \times 10^{- 19} - 2 .229 \times 10^{- 19} \\ = 2 .401 \times 10^{- 19} J \\ = \frac{2 .401 \times 10^{- 19}}{1 .6 \times 10^{- 19}} = 1 .5 eV \end{aligned}$

Hence, the new stopping potential is 1.50 eV.

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