Question

A metal surface is illuminated by light of two different wavelengths $248\,nm$ and $310\,nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are ${u_1}$ and ${u_2}$, respectively. If the ratio ${u_1}:{u_2} = 2:1$ and $hc = 1240eV\,nm$, the work function of the metal is nearly:
(A) $3.7eV$
(B) $3.2eV$
(C) $2.8eV$
(D) $2.5eV$

Answer 1

Hint: On the basis of Planck’s quantum theory, Albert Einstein found an equation which defines the photoelectric effect which is commonly known as Einstein photoelectric equation. By using this Einstein photoelectric equation, the work function of the metal is determined.

Useful formula
Einstein photoelectric equation,
$K.E = h\nu - {\phi _0}$
Where $K.E$ is the Kinetic energy, $h$ is the Planck constant, $\nu $ is the frequency of radiation, ${\phi _0}$ is the work function

Complete step by step solution
By using Einstein photoelectric equation,
$K.E = h\nu - {\phi _0}\,...................\left( 1 \right)$
Here $K.E$ is the kinetic energy, $h\nu $ is the energy of photon, ${\phi _0}$ is the work function
As we know the Kinetic energy formula,
$K.E = \dfrac{1}{2} \times m{v^2}$
Substitute the kinetic energy formula in equation (1),
$\dfrac{1}{2} \times m{v^2} = h\nu - {\phi _0}\,...............\left( 2 \right)$
Where $m$ is the mass of the object, $v$ is the velocity of the object
The frequency of radiation,
 $\nu = \dfrac{c}{\lambda }$
Substitute the threshold frequency value in the equation (2),
$\dfrac{1}{2} \times m{v^2} = \dfrac{{hc}}{\lambda } - {\phi _0}\,...............\left( 3 \right)$
The metal surface is illuminated by light of two different waves, then,
$\dfrac{1}{2} \times m{v_1}^2 = \dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}$ and $\dfrac{1}{2} \times m{v_2}^2 = \dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}$
Hence the velocity is given as $u$ in the question, then the above equations are changed as,
$\dfrac{1}{2} \times m{u_1}^2 = \dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}$ and $\dfrac{1}{2} \times m{u_2}^2 = \dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}$

By taking the ratio of two equations,
$\dfrac{{\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}}}{{\dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}}} = \dfrac{{\dfrac{1}{2} \times m{u_1}^2}}{{\dfrac{1}{2} \times m{u_2}^2}}$
Cancelling the same term in RHS,
$\dfrac{{\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}}}{{\dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}}} = \dfrac{{{u_1}^2}}{{{u_2}^2}}$
Taking the square as common in the above equation in RHS,
$\dfrac{{\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}}}{{\dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}}} = {\left( {\dfrac{{{u_1}}}{{{u_2}}}} \right)^2}$
Substituting the velocity ratio in the above equation in RHS,
$\dfrac{{\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}}}{{\dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}}} = {\left( {\dfrac{2}{1}} \right)^2}$
By solving the above equation in RHS,
$\dfrac{{\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0}}}{{\dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}}} = 4$
By rearranging the above equation,
$\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0} = 4\left( {\dfrac{{hc}}{{{\lambda _2}}} - {\phi _0}} \right)$

By multiplying the values in RHS,
$\dfrac{{hc}}{{{\lambda _1}}} - {\phi _0} = \dfrac{{4hc}}{{{\lambda _2}}} - 4{\phi _0}$
Taking the work function value in one side and other values in other side,
$4{\phi _0} - {\phi _0} = \dfrac{{4hc}}{{{\lambda _2}}} - \dfrac{{hc}}{{{\lambda _1}}}$
By subtracting the work function in LHS,
$3{\phi _0} = \dfrac{{4hc}}{{{\lambda _2}}} - \dfrac{{hc}}{{{\lambda _1}}}\,....................\left( 4 \right)$
Substituting the $hc$, ${\lambda _1}$ and ${\lambda _2}$ in the equation (4),
$3{\phi _0} = \left( {\dfrac{{4 \times 1240}}{{310}}} \right) - \dfrac{{1240}}{{248}}$
By solving the above equation in RHS,
$
  3{\phi _0} = 16 - 5 \\
  3{\phi _0} = 11 \\
 $
By solving the above equation,
$
  {\phi _0} = \dfrac{{11}}{3} \\
  {\phi _0} = 3.66eV \\
  {\phi _0} \simeq 3.7eV \\
 $

Hence, the option (A) is correct.

Note: The incident photons from the source which collides with the electron inside an atom and emits the energy of the electron. A part of this energy is used by the electron to escape from the surface of the metal and the remaining part will be the kinetic energy with which the electron is ejected.