Question

Photoelectric threshold wavelength of metallic silver is $\lambda -3800A$ . Ultra violet light of $\lambda =2600A$ is incident on the silver surface. Calculate
(a) the value of work function in joule and $eV$
(b) maximum kinetic energy of the emitted photoelectrons,
(c) the maximum velocity of the photo electrons. (Mass of the electron $=9.11\times {10}^{-31}Kg$ ).

Answer 1

Hint: This problem is based on the concept of the movement of the photons which moves with the certain kinetic energy and the velocity. Work function is defined as the minimum amount of the energy required to remove the electron from the surface to the position of the infinity. The maximum kinetic energy is obtained by subtracting the total energy by work function.

Formula used:
(1) The formula of the work function is given by
$w={\displaystyle \frac{hc}{{\lambda }_0}}$
Where $w$ is the work function, $h$ is the Planck’s constant, $c$ is the speed of the light and ${\lambda }_0$ is the wavelength of the silver.
(2) The formula for the energy is given as
$E={\displaystyle \frac{hc}{\lambda }}$
Where $\lambda$ is the wavelength of ultraviolet light
(3) The formula for the maximum kinetic energy sis given by is given as
$K{E}_{max}=E-w$
$K{E}_{max}={\displaystyle \frac{1}{2}}m{v_{max}}^2$
Where $E$ is the energy of the electrons, $K{E}_{max}$ is the maximum kinetic energy, $m$ is the mass of the electron and $v_{max}$ is the maximum velocity.

Complete step by step answer:
Given: Threshold wavelength of metallic silver, ${\lambda }_0=3800A$
The wavelength of the ultra violet light, $\lambda =2600A$
(a) Using the formula of the work function,
$w={\displaystyle \frac{hc}{{\lambda }_0}}$
Substituting the known values in it, we get
$w={\displaystyle \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{3800\times {10}^{-10}}}$
By simplifying the above equation,
$w=5.23\times {10}^{-19}J$
$w=3.27eV$
(b) By using the formula of the maximum kinetic energy,
$K{E}_{max}=E-w$
Substituting the formula of energy in it,
$K{E}_{max}=E-w$
$K{E}_{max}={\displaystyle \frac{hc}{\lambda }}-w$
Substituting the values in it, we get
$KE={\displaystyle \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{2600\times {10}^{-10}\times 1.6\times {10}^{-19}}}-3.27$
By simplification, we get
$K{E}_{max}=1.51eV$
(c) By substituting the formula of kinetic energy in above equation,
${\displaystyle \frac{1}{2}}m{v_{max}}^2=1.51eV$
By substituting the value of the mass of electron in it,
$v_{max}=\sqrt{\left({\displaystyle \frac{1.51\times 2}{9.1\times {10}^{-19}}}\right)}eV$
By simplifying, we get
$v_{max}=0.789\times {10}^{6}m{s}^{-1}$

Hence, the maximum velocity of the electron is $0.789\times {10}^{6}m{s}^{-1}$.

Note: The value of the Planck’s constant is taken as $6.63\times {10}^{-34}$ and the value of the speed of light is taken as $3\times {10}^8$ . The joule is converted into electron volt by dividing it with the charge of the electron ($1.6\times {10}^{-19}$ ).