Question

In a photoelectric effect experiment, electrons are ejected from a titanium surface (work function, 3 eV) following irradiation with UV light. The energy of the incident UV light is 7.2 x 10-19 J.
(a) Calculate the wavelength of the ejected electrons.
(b) Calculate the wavelength of the incident UV light.
(c) would an iron surface (Ф-4.7 eV require a longer or shorter wavelength of light to eject electrons with the same wavelength calculated in part (a)? Briefly explain.

Answer 1

Explanation:

According to the Einstein law, it is known that

            $h \times \nu = \phi + \frac{1}{2} mv^{2}$

where,   h = energy of light

           $\phi$ = work function

           $m v^{2}$ = kinetic energy of electron

It is given that the value of $h \nu$ is $7.2 \times 10^{-19} J$. And,

               1 eV = $1.6 \times 10^{-19} J$

Here,  $\phi$ for titanium is 4.33 eV

          = $(4.33 \times 1.6 \times 10^{-19})$ J  

          = $6.928 \times 10^{-19}$ J

(a)   First of all, kinetic energy will be calculated as follows.

             $\frac{1}{2}mv^{2} = h \nu - \phi$

                         = $(7.2 \times 10^{-19} - 6.92 \times 10^{-19})$ J

                         = $0.272 \times 10^{-19}$ J

It is known that mass of electrons is equal to $9.109 \times 10^{-31} kg$.

Therefore, $mv^{2} = 0.544 \times 10^{-19} J$

and,     $(mv)^{2} = 9.109 \times 0.544 \times 10^{-19} \times 10^{-31}$

                       = $4.955 \times 10^{-50}$

                mv = $2.225 \times 10^{-25}$

Now, the relation between wavelength and mv is as follows.

      $\lambda = \frac{6.626 \times 10^{-34}}{2.225 \times 10^{-25}}$

                   = $2.98 \times 10^{-9} m$

Therefore, the wavelength of the ejected electrons is $2.98 \times 10^{-9} m$.

(b)   It is known that relation between energy and wavelength is as follows.

               E = $h \nu = \frac{hc}{\lambda}$

                  $\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{7.2 \times 10^{-19}}$

                     = $\frac{6.626 \times 3 \times 10^{-26}}{7.2 \times 10^{-19}}$

                    = $2.76 \times 10^{-7} m$

Hence, the wavelength of the ejected electrons is $2.76 \times 10^{-7} m$.

(c)   For iron surface, $\phi = 4.7 eV$

                                       = $(4.7 \times 1.6 \times 10^{-19})$ J

                                       = $7.52 \times 10^{-19}$ J

Here, the value of $\phi$ is more than the value of UV light source. Hence, we need a shorter wavelength light as we know that,

                  $E \propto \frac{1}{\lambda}$

Therefore, lesser will be the wavelength higher will be the energy.