In fig, electromagnetic radiations of wavelength $200 n m$ are incident on a metallic plate A. The photoelectrons are accelerated by a potential difference of $10 V$ .These electrons strike another metal plate B from which electromagnetic radiations are emitted. The minimum wavelength of emitted photons is $100 n m$ . If the work function of metal A is found to be $(x \times 10^{- 1}) e V$ , then find the value of $x$ . (Given $h c = 1240 e V n m)$

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Solution

Let us first find the initial energy of the electron just after the light strikes the plate :
$E_{i}$ = $\frac{h C}{λ}$ - $ϕ$ = $\frac{1240}{200} - ϕ$ = $6.2 - ϕ$
After this it gains energy from the battery. Now the total energy becomes :
$E_{f}$ = $E_{i}$ + $10$ = $16.2 - ϕ$
This energy is released as light from the other plate .
$\frac{h C}{λ_{2}}$ = $16.2 - ϕ$
$\frac{1240}{100}$ = $16.2 - ϕ$
$ϕ = 16.2 - 12.4 = 3.8 = 38 \times 10^{- 1}$
Hence $x = 38$

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