Question

The work function for the cesium atom is $1.9\,eV$. Calculate:
(a) the threshold wavelength and
(b) the threshold frequency of the radiation. If the cesium element is irradiated with a wavelength of $500\,nm$, calculate the kinetic energy and the velocity of the photoelectron.

Answer 1

Einstein’s photoelectric equation states that the energy of a photon hitting the surface of a metal is the sum of the work function of the metal and the energy of the released electron from the metal.

Formula used:
Einstein’s equation for photoelectric effect is given by,
${E_p} = W + {E_e}$
where, ${E_p}$ is the energy of the photon, $W$ is the work function of the metal and ${E_e}$ is the kinetic energy of the electron.
Energy of a photon is given by,
$E = h\nu$
where, $h$ is the Planck’s constant and $\nu$ is the frequency of the photon.
Relation between wavelength and frequency is given by,
$\lambda = \dfrac{c}{\nu }$
where,$c$ is the velocity of light and $\nu$ is the wavelength.
Kinetic energy of any particle is given by,
$E = \dfrac{1}{2}m{v^2}$
where, $m$ is the mass of the particle and $v$ is the velocity of the particle

Complete step by step answer:
We know from photoelectric effect, when a photon hits the surface of any metal a photoelectron comes out. The minimum energy required to release the electron is known as the work function of that metal. Einstein’s equation for the photoelectric effect states that, energy of a photon hitting any metal surface is the sum of the work function of the metal and the energy of the released electron. It is given by, ${E_p} = W + {E_e}$. Now, we have given here the work function of cesium is, $W = 1.9eV$.

(a) Now, for threshold energy the work function of the metal is just equal to the energy of the photon. So, at threshold condition we can write, ${E_p} = W$. Now, energy of a photon is given by, $E = h\nu$ where, $h$ is the Planck’s constant and $\nu$ is the frequency of the photon. Hence, we can write, $W = h\nu$.
So, putting, $h = 6.62 \times {10^{ - 34}}$ and $W = 1.9eV$ and $1eV = 1.602 \times {10^{ - 19}}J$ we have,
$6.62 \times {10^{ - 34}}\nu = 1.9 \times 1.602 \times {10^{ - 19}}$
Solving we have,
$\nu = \dfrac{{3.0438 \times {{10}^{15}}}}{{6.62}}$
$\Rightarrow \nu = 4.598 \times {10^{14}}$

Now, relation between wavelength and frequency is given by,
$\lambda = \dfrac{c}{\nu }$
Hence, threshold wavelength becomes,
$\lambda = \dfrac{{3 \times {{10}^8}}}{{4.598 \times {{10}^{14}}}}$
$\therefore \lambda = 6.538 \times {10^{ - 7}}$
Hence, threshold wavelength is $6.538 \times {10^{ - 7}}m$ or $653.8nm$.

(b) From the above calculation we can write the threshold frequency is $4.598 \times {10^{14}}\,Hz$.

(c) Now, for a photon with $\lambda = 500\,nm = 5 \times {10^{ - 7}}\,m$ energy of the photon is,
${E_p} = \dfrac{{hc}}{\lambda }$
Putting the value we have,
${E_p} = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5 \times {{10}^{ - 7}}}}J$
$\Rightarrow {E_p} = 3.972 \times {10^{ - 19}}J$ [Putting, $1eV = 1.602 \times {10^{ - 19}}J$]
$\Rightarrow{E_p} = 2.48eV$
So, putting the values of ${E_p} = 2.48eV$ and $W = 1.9eV$ in Einstein’s equation we have,
$2.48 = 1.9 + {E_e}$
So, it becomes,
$\therefore {E_e} = 2.48 - 1.9 = 0.58$
So, the kinetic energy of the electron is $0.58\,eV$.

Now, we know kinetic energy,
$E = \dfrac{1}{2}m{v^2}$
So, the velocity of the electron will be,
$v = \sqrt {\dfrac{{2{E_e}}}{m}}$
Putting the mass of electron $m = 9.11 \times {10^{ - 31}}kg$and ${E_e} = 0.58eV = 3.972 \times {10^{ - 19}}J$ we have,
$v = \sqrt {\dfrac{{2 \times 3.972 \times {{10}^{ - 19}}}}{{9.11 \times {{10}^{ - 31}}}}}$
Calculating we have,
$\therefore v = 4.521 \times {10^6}$
So, the velocity of the photoelectron will be, $4.521 \times {10^6}m{s^{ - 1}}$.

Note: Work function of any metal surface is the minimum energy to just release from the metal surface that means the electron will just come out of the influence of the nucleus and will be at rest after leaving the metal. Work function for different metals is different since the ionization energy of every element is different from the other. If metal is kept as a terminal of an electrical circuit we will observe a voltage drop across the terminal when the photon hits the surface and the kinetic energy will be then equal to ${E_e} = eV$ where $e$ is the charge of electron and $V$ is the voltage drop across the terminal.