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Photoelectric Effect and WaveParticle Duality — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsPhotoelectric Effect and WaveParticle Duality5 Marks
Question
The electric field associated with a light wave is given by $\text{E}=\text{E}_0\sin[(1.57\times10^7\text{m}^{-1})(\text{x}-{\text{ct}})].$ Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9eV.
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Answer

$\text{E}=\text{E}_0\sin[(1.57\times10^{7}\text{m}^{-1})(\text{x}-\text{ct})]$$\text{w}=1.57\times10^7\times\text{c}$
$\Rightarrow\text{f}=\frac{1.57\times10^{7}\times3\times10^{8}}{\text{2}\pi}\text{Hz}$
$\text{w}_0=1.9\text{ev}$
Now $eV_0 = hv - W_0$
$=4.14\times10^{-15}\times\frac{1.57\times3\times10^{15}}{2\pi}-1.9\text{ev}$
$=3.105-1.9=1.205\text{ev}$
So, $\text{v}_0=\frac{1.205\times1.6\times10^{-19}}{1.6\times10^{-19}}=1.205\text{ev.}$

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