Question
In the arrangement shown in figure $y = 1.0mm. d = 0.24mm$ end $D = 1.2m$. The work function of the materiel of the emitter is $2.2eV$. Find the stopping potential V needed to stop the photocurrent.
✓
Answer
Given : fringe width, $y = 1.0mm × 2 = 2.0mm, D = 0.24mm, W_0 = 2.2ev, D = 1.2m \text{y}=\frac{\lambda\text{D}}{\text{d}}$
$\lambda=\frac{\text{yd}}{\text{D}}=\frac{2\times10^{-3}\times0.24\times10^{-3}}{1.2}=4\times10^{-7}\text{m}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{4.14\times10^{-15}\times3\times10^{8}}{4\times10}=3.105\text{ev.}$
Stopping potential $eV_0 = 3.105 - 2.2 = 0.905V$
$\lambda=\frac{\text{yd}}{\text{D}}=\frac{2\times10^{-3}\times0.24\times10^{-3}}{1.2}=4\times10^{-7}\text{m}$
$\text{E}=\frac{\text{hc}}{\lambda}=\frac{4.14\times10^{-15}\times3\times10^{8}}{4\times10}=3.105\text{ev.}$
Stopping potential $eV_0 = 3.105 - 2.2 = 0.905V$
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