Question

In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon (h) absorbed by a metal is the sum of the work function (), the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy (Ek) of the electron: h = + Ek. When light of wavelength 357.4 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 5.8×105 m/s. What is (in kJ/mol) of potassium?

Answer 1

Complete Question

 In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon (h) absorbed by a metal is the sum of the work function ($\phi$), the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy (Ek) of the electron: $hf = \phi +Ek$. When light of wavelength 357.4 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 5.8×105 m/s. What is work function (in kJ/mol) of potassium?

Answer:

The work function is  $\phi_m = 242.195 \ KJ /mol$

Explanation:

From the question we are told that

     $hf = \phi +Ek$

   Where h is the Planck's constant which has  a constant value of  $h = 6.626 *10^{-34} J \cdot s$

              Ek  is the kinetic energy which is mathematically represented as

                 $Ek = \frac{1}{2} m v^2$

Where m  is the mass of electron with a constant value  

               $m = 9.11 *10^{-31} \ kg$

     The wavelength is  $\lambda = 357.4 nm = 357.4 *10^{-9} \ m$

     The speed of the electron is $v = 5.8*10^{5} m/s$

This photoelectric effect  can be mathematically defined as

f is the frequency of the incident light  which is mathematically represented as  $f = \frac{c}{\lambda }$

   $\phi$ is the work function

So  we have  

             $\frac{hc}{\lambda } = \phi + \frac{1}{2} mv^2$

substituting value

              $\frac{6.626 *10^{-34}* 3*10^8}{357.4 *10^{-9}} = \phi + \frac{1}{2} * (9.11*10^{-31}) * (5.8*10^{8}) ^2$

              $5.5618*10^{-19} = \phi + 1.54*10^{-14}$

      =>     $\phi = 4.021*10^{-19} J$

For one mole of  one mole the work function is  

        $\phi _m = \phi * N$

Where N is the avogadro's constant with a value  

             $N = 6.022*10^{23}$

So

       $\phi_m = 4.021*10^{-19} * 6.022 *10^{23}$

       $\phi_m = 242194.90 J /mol$

now in KJ

        $\phi_m = 242.195 \ KJ /mol$