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The threshold frequency ${{v}_{0}}$ for a metal is$7.0\times{{10}^{14}}{{s}^{-1}}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $v=1.0\times{{10}^{15}}{{s}^{-1}}$ hits the metal.

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Last updated date: 15th Jun 2024
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Answer
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Hint: The electrons are ejected from the surface of metal as soon as the beam of light strikes the surface. For each metal certain minimum frequency is needed to eject the electrons. This is known as threshold frequency and is different for different metals. According to Einstein when a photon strikes the surface the energy of striking photon is expressed as:
Energy of striking photon = Binding energy + Kinetic energy of ejected electron.

Formula used: $hv=h{{v}^{\circ }}+\dfrac{1}{2}m{{v}^{2}}$
Where, $h$ is the planck's constant
$v$ is the frequency of photon
${{v}^{\circ }}$ is the threshold frequency
$m$ is the mass

Complete step by step answer:
According to Einstein when a photon strikes a metal surface, some of its energy is used up to eject electrons from the metal atom and the remaining energy is used to eject electrons in the form of kinetic energy. Some of the energy of striking photons called threshold energy $x$ is used to remove electrons from the surface and the remaining energy is imparted to the ejected electron as kinetic energy.
Therefore;

$hv=x+\dfrac{1}{2}m{{v}^{2}}$

If the threshold frequency is ${{v}_{0}}$ then the threshold energy $x=h{{v}_{0}}$ so that:

$\Rightarrow hv=h{{v}_{0}}+\dfrac{1}{2}m{{v}^{2}}$

$\Rightarrow \dfrac{1}{2}m{{v}^{2}}=hv-h{{v}_{0}}$
This is equation $1$

In the question it is given that the threshold frequency ${{v}_{0}}$=$7\times{{10}^{14}}{{s}^{-1}}$, v=$1.0x{{10}^{15}}{{s}^{-1}}$ and h= $6.626x{{10}^{-34}}{Js}$. Putting the values in equation 1 we get the value of kinetic energy:

$\Rightarrow 6.626x{{10}^{-34}}(10x{{10}^{14}}-7x{{10}^{14}}) \\ $

$\Rightarrow 6.626\times {{10}^{-34}}\left( 3\times {{10}^{14}} \right)$

$\Rightarrow 19.878\times {{10}^{-20}}J$

Hence, the kinetic energy of the ejected electron is $19.8\times{{10}^{-20}}J$ .

Note: The number of photoelectrons ejected per second from the metal depends upon the intensity or brightness of incident radiation but does not depend upon its frequency. The kinetic energy of the photoelectron remains unchanged on increasing the intensity of radiation and increases with the frequency of incident radiation.