Threshold frequency, v_0 is the minimum frequency which a photon …

Question

Threshold frequency, $v_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0 \times 10^{15} \mathrm{~s}^{-1}$ was allowed to hit a metal surface, an electron having $1.988 \times 10^{-19}$ J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

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Answer

$\text{hv}=\text{h}_0+\text{K.E}.$
$\text{hv}_0=\text{hv}-\text{K.E.}$
$\text{v}_0=\text{v}-\frac{\text{K.E.}}{\text{h}}\ ...(\text{i})$
$\text{v}=1.0\times10^{15}\text{s}^{-1}$
$\text{K.E.}=1.988\times^{-19}\text{J},\text{ h}=6.626\times10^{-34}\text{Js}$
From (i) we have,
$\text{v}_0=1.0\times10^{15}\text{s}^{-1}-\frac{1.988\times10^{-19}\text{J}}{6.626\times10^{-34}\text{Js}}$
$=(1.0\times10^{15}-0.30\times10^{15})\text{s}^{-1}$
$=0.7\times10^{15}\text{s}^{-1}=7\times10^{14}\text{s}^{-1}$
$\lambda=600\text{nm}=600\times10^{-9}\text{m}=6.0\times^{-7}\text{m}$
$\text{v}=\frac{\text{v}}{\lambda}=\frac{3.0\times10^8\text{ms}^{-1}}{6.0\times10^{-1}\text{m}}=5\times10^{14}\text{s}^{-1}$
Thus $\text{v}$ < $\text{v}_0$, Hence, no electron will be emitted.

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