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 2.0 eV.

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Solution

The electric field becomes $0$ , $1.2 \times 10^{15}$ times per second.
$∴$ Frequency = $\frac{1.2 \times 10^{15}}{2} = 0.6 \times 10^{15}$ Hz
hv = $ϕ_{0} + K . E$
$\Rightarrow h v - ϕ_{0} = K . E$
$\Rightarrow K E = \frac{6.63 \times 10^{- 34} \times 0.6 \times 10^{15}}{1.6 \times 10^{- 19}} - 2$
$K . E = 0.482$ $e v$ $=$ $0.48$ $e v$ .

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The electric field associated with a monochromatic beam becomes zero, 1.2×1015 times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

The electric field becomes 0, 1.2×1015 times per second.∴ Frequency = 1.2×10152=0.6×1015 Hzhv = ϕ0+K.E⇒hv−ϕ0=K.E⇒KE=6.63×10−34×0.6×10151.6×10−19−2K.E=0.482 ev = 0.48 ev.

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 2.0 eV.