Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^–9 m) from a neon lamp irradiates a photosensitive material made of caesium 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 photocell. Predict the new stopping voltage.

Solution

Here, we are given a neon lamp.

W a v e l e n g t h o f r a d i a t i o n, λ = 640.2 n m = 640.2 \times 10^{- 9} m S t o p p i n g p o t e n t i a l, V_{1} = 0.54 V

The source is replaced by an iron source whose wavelength is 427.2 nm.
Stopping potential applied to the iron source, V₂ = ?

Using the formula of Einstein's energy equation,

{eV}_{1} = \frac{hc}{λ_{1}} - ϕ_{0}

{eV}_{2} = \frac{hc}{λ_{2}} - ϕ

On subtracting both equations,

{eV}_{2} - {eV}_{1} = hc [\frac{1}{λ_{2}} - \frac{1}{λ_{1}}]

\Rightarrow

V_{2} - V_{1} = \frac{hc}{e} [\frac{1}{λ_{2}} - \frac{1}{λ_{1}}]

\Rightarrow

V_{2} = V_{1} = \frac{hc}{e} [\frac{1}{λ_{2}} - \frac{1}{λ_{1}}]

\Rightarrow

V_{2} = 0.54 + \frac{6.63 \times 10^{- 24} \times 3 \times 10^{8}}{1.6 \times 10^{- 19}} [\frac{1}{427.2 \times 10^{- 9}} - \frac{1}{640.2 \times 10^{- 9}}]

= 0.54 + \frac{6.63 \times 3 \times 10^{- 24 + 8 + 9}}{1.6} [\frac{1}{427.2} - \frac{1}{640.2}]

= 0.54 + \frac{6.63 \times 3}{1.6} \frac{[640.2 - 427.2]}{640.2 \times 427.2} \times 10^{- 7}

= 0.54 + 0.97

That is ,

V_{2} = 1.51 V .

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