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# Energy required to stop the ejection of electrons from $Cu\;$ plate is $0.24\;eV$.When radiation of $\lambda=253.7\;nm$ strikes the plate. Calculate the work function.

Last updated date: 13th Jul 2024
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Hint:To solve the question, we need to understand the concept of photoelectric effect. To get the ejection of electrons, the incident energy must be enough to overcome the work function. Here, we are given the energy to stop the emission of electrons, which will be equal to the resultant energy with which the electrons were emitted.

Formula used:
${{E}_{in}}=\dfrac{hc}{\lambda }$
Where, $E_{in}$ = Incident energy
$h$ = Planck’s constant = $6.626\times {{10}^{-34}}{{m}^{2}}kg{{s}^{-1}}$
$c$ = Speed of light in vacuum = $3\times {{10}^{8}}m{{s}^{-1}}$

Let us first understand the concept of photoelectric effect. We know that metals have free electrons which can be obtained as emission of electrons. However, these electrons cannot come out of the metallic surface on their own. The reason for this is that the electrons are attracted inwards by the nucleus, while there is no force pulling them outside.Hence, we can say a potential barrier exists at the surface.

Thus, the minimum energy required to just pull out the electrons to the surface of the metal or the minimum energy required to cross the potential barrier is known as work function.Hence, to get a complete emission of electrons, we need to apply the energy more than the work function in the form of bombardment of photons. We can understand from the above definition that the net energy possessed by the emitted electrons will be the difference between the supplied energy and the work function.
${{\text{W}}_{\text{net}}}\text{=Incident Energy - Work function}$

This will be the net energy of the electrons that are emitted. So if we need to stop the emission of the electrons we need to provide the equal amount of energy.
$\text{Energy required to stop the emission = Incident Energy - Work function}$
Here, we need to find the work function which can be represented as $\phi$
$\therefore \text{Energy required to stop the emission = Incident Energy - }\phi$ . . . . . . . . $(1)$
Now, we know that the incident energy is due to the bombardment of photons and can be calculated as,
${{E}_{in}}=\dfrac{hc}{\lambda }$
$\lambda$ = Wavelength of the radiation of photons = $253.7\times {{10}^{-9}}m$ (given)

Substituting all the above given values, incident energy is obtained as
${{E}_{in}}=\dfrac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}}{253.7\times {{10}^{-9}}}$
$\Rightarrow {{E}_{in}}=0.07835\times {{10}^{-17}}\,J$
Now, we are given the energy in units of electron volt $(eV)$ . Hence, to convert the above answer to required unit we need to divide the answer by $1.6\times {{10}^{-19}}$
${{E}_{in}}=\dfrac{0.07835\times {{10}^{-17}}}{1.6\times {{10}^{-19}}}$
$\therefore {{E}_{in}}=4.897\,eV$
Substituting the above value and the given values in the equation $(1)$,
$0.24\,eV=4.897\,eV-\phi$
$\therefore \phi =4.657\,eV$

Hence, the work function is $4.657\;eV$.

Note: In this question and similar type of questions, we need to note the conversion value of Joules into Electron volts or vice-versa i.e. $1J=1.6\times {{10}^{-19}}eV$. We should remember to convert all the values to the same unit. The unit of electrons volt is used to represent very small values of energy.