Question
The electric field associated with a monochromatic beam becomes zero $1.2 \times 10^{15}$ times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0eV.
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Answer
The electric field becomes $01.2 \times 10^{45}$ times per second.
$\therefore$ Frequency $=\frac{1.2\times10^{15}}{2}=0.6\times10^{15}$
$\text{hv}=\phi_0+\text{kE}$
$\Rightarrow\text{hv}-\phi_0=\text{KE}$
$\Rightarrow\text{KE}=\frac{6.63\times10^{-34}\times0.6\times10^{15}}{1.6\times10^{-19}}-2$
$=0.482\text{ev}=0.48\text{ev.}$
$\therefore$ Frequency $=\frac{1.2\times10^{15}}{2}=0.6\times10^{15}$
$\text{hv}=\phi_0+\text{kE}$
$\Rightarrow\text{hv}-\phi_0=\text{KE}$
$\Rightarrow\text{KE}=\frac{6.63\times10^{-34}\times0.6\times10^{15}}{1.6\times10^{-19}}-2$
$=0.482\text{ev}=0.48\text{ev.}$
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