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Photoelectric Effect and WaveParticle Duality — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPhotoelectric Effect and WaveParticle Duality5 Marks
Question
The electric field at a point associated with a light wave is $\text{E}=\big(100\frac{\text{v}}{\text{m}}\big)\sin[(3.0\times10^{15}\text{s}^{-1})\text{t]sin[(6.0}\times10^{15}\text{s}^{-1})\text{t}].$ If this light falls on a metal surface having a work function of $2.0eV$, what will be the maximum kinetic energy of the photoelectrons?
✓

Answer

$\text{E}=100\sin[(3\times10^{15}\text{s}^{-1})\text{t}]-\cos[3\times10^{15}\text{s}^{-1} )\text{t]}$$=100\frac{1}{2}[\cos[9\times10^{15}\text{s}^{-1})\text{t}]-\cos[3\times10^{15}\text{s}^{-1})\text{t}]$
The w are $9 \times 10^{15}$ and $3 \times 10^{15}$ for largest K.E.
$\text{f}_\text{max}=\frac{\text{W}_{\text{max}}}{2\pi}=\frac{9\times10^{15}}{2\pi}$
$\text{E}-\phi=\text{K.E}.$
$\Rightarrow\text{hf}-\phi_0=\text{K.E}.$
$\Rightarrow\frac{6.63\times10^{-34}\times9\times10^{15}}{2\pi\times1.6\times10^{-19}}-2=\text{KE}$
$\Rightarrow\text{KE}=3.938\text{ev}=3.93\text{ev}.$

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