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What is the wave function of the metal if the light of wavelength 4000${{A}^{{}^\circ }}$generates photoelectrons of velocity $6\times {{10}^{5}}m{{s}^{-1}}$?

 [Given: Mass of electron = $9\times {{10}^{-31}}Kg$
 Velocity of light = $3\times {{10}^{8}}m{{s}^{-1}}$
 Planck's constant = $6.626\times {{10}^{-34}}Js$
 Charge of electron= $1.6\times {{10}^{-19}}Jev$]

A. 0.9 eV
B. 4.0 eV
C. 2.1 eV
D. 3.1eV


Answer
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Hint: To find the work function we have a formula of $K.E=\phi -E$ where K.E is the kinetic energy of electron, E is the energy of the incoming photon and $\phi $is the work function many places it is also termed as wave function. Energy of a incoming photon is calculated by the formula $E=\dfrac{h}{\lambda }C$ which can be directly written as $E=\dfrac{12400}{\lambda ({{A}^{{}^\circ }})}$ where $\lambda $ is the wavelength.

Complete step by step solution:
 From your chemistry lessons you have learned about the Kinetic energy of photoelectrons, their wave function and the Einstein equation to find the energy of photons. Wave function is defined as the minimum amount of energy required to remove one electron from the metal surface but it is important to know that all the electrons that are removed from the metal surface will not have the same kinetic energy. Each electrons will be removed with different kinetic energy and this happens because the photons that are required to remove electron carries energy and all the energy of photons will not be transferred to the metal surface, some of the energy is transferred to the electrons that are not present near the surface or we can say that it is transferred to the bulk of the metal. Therefore the electrons which emit fast from the metal will absorb all the energy of the incoming photons which is not needed to remove the electron from the metal surface and thus the kinetic energy of all the electrons will not remain the same. The maximum kinetic energy of the emitted electron is expressed as
\[K.{{E}_{\max }}=\phi -{{E}_{photon}}\]……………….. (1)
Where $\phi $ is the wave function/work function of the metal
\[{{E}_{photon}}\]= energy of the photon
As we know that to calculate kinetic energy of we have a formula that is,
\[K.E=\dfrac{1}{2}m{{v}^{2}}\]
Where, m= mass of the electron,
V= velocity of electron
Now to calculate the energy of photon we use the Planck- Einstein equation, which is given as
\[E=\dfrac{h}{C}\lambda \]…………………. (2)
Where, h= Planck's constant which is equal to $6.626\times {{10}^{-34}}Js$
C= speed of light in vacuum given as $3\times {{10}^{8}}m{{s}^{-1}}$
$\lambda $= wavelength of the photon
-Now if you will put the value of h and C in the formula you will get Energy as,
\[E=\dfrac{12400}{\lambda ({{A}^{{}^\circ }})}\]……………………. (3)
-Now put the values of (2) and (3) in equation (1), you will get
$\dfrac{1}{2}m{{v}^{2}}=\phi -\dfrac{12400}{\lambda ({{A}^{{}^\circ }})}$………….. (4)
-Now , in the question the value of velocity of electron, mass of electron and the wave length is given so put them in equation (4),
\[\dfrac{1}{2}\times 9.1\times {{10}^{-31}}\times {{(6\times {{10}^{5}})}^{2}}=\phi -\dfrac{12400}{4000{{A}^{{}^\circ }}}\]
\[1\,ev=\phi -3.1\,ev\]
\[\phi =2.1\,ev\]

Thus the correct option will be (C).

Note: For the emission of electrons the work function should be less than the energy of photons if the work function will be greater than energy of photon then the incoming photons will have enough energy, And thus it will not overcome the work function and will not emit the electrons from the metal surface. To convert the value from Joule into eV we have, $1Joule\,=\,6.24\times {{10}^{18}}eV$.