Question

When a metal surface is illuminated by light of wavelengths $400nm$ and $250nm$, the maximum velocities of the photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is
A. \[2hc \times {10^6}J\]
B. \[1.5hc \times {10^6}J\]
C. \[hc \times {10^6}J\]
D. \[0.5hc \times {10^6}J\]

Answer 1

Hint: The equation for maximum kinetic energy of the photoelectrons ejected as a result of photoelectric emission is given as $KE = E - W$, $KE$ is the kinetic energy of the photoelectrons $\left( {KE = \dfrac{1}{2}m{v^2}} \right)$, $W$ is the work function of the metals surface and $E$ is energy of the incident light$\left( {E = \dfrac{{hc}}{\lambda }} \right)$.Upon substituting the values given in the question and solving the question we get the value of work function.

Complete step by step answer:
We know that the energy of light can be written as $E = \dfrac{{hc}}{\lambda }$.
where $h$ is Planck's constant, $c$ is the speed of light and $\lambda $ is the wavelength of light.
The equation for maximum kinetic energy of photoelectrons emitted as a result of photoelectric emission is $KE = E - W$,
We know that the kinetic energy is written as $\left( {KE = \dfrac{1}{2}m{v^2}} \right)$,
where $m$ is the mass and $v$ is the velocity of the photoelectron.
Therefore the equation becomes,
$\dfrac{1}{2}m{v^2} = \dfrac{{hc}}{\lambda } - W$ -----------(1)
Given when the wavelength of light is 400nm then the maximum velocity of photoelectron is V. Substituting these values in the equation 1 we get,
$\dfrac{1}{2}m{V^2} = \dfrac{{hc}}{{400}} - W$ -------------(2)

When the wavelength of light is 250nm then the maximum velocity of the photoelectron is 2V. Substituting these values in the equation 1 we get,
$\dfrac{1}{2}m4{V^2} = \dfrac{{hc}}{{250}} - W$ -------------(3)
Multiplying to 4 equation 2
$\dfrac{1}{2}m4{V^2} = \dfrac{{hc}}{{100}} - 4W$ --------------(4)
Subtracting equation 4 from 3 we get
$0 = - \dfrac{{3hc}}{{(500)nm}} + 3W \\
\Rightarrow 3W = \dfrac{{3hc}}{{(500)nm}} \\
\Rightarrow W = \dfrac{{hc}}{{500}} \times {10^9}\left( {1nm = {{10}^{ - 9}}m} \right) \\
\therefore W = 2hc \times {10^6}J \\ $

Hence, the work function of the metal is $2hc \times {10^6}J$.

Note: We know that metal has free electrons (negatively charged particles) in its surface. But free electrons cannot escape from the metal surface unless there is a push or pull. If electrons attempt to escape from the surface of the metal, it will acquire a positive charge hence it will attract back the electrons to the surface. So as mentioned earlier it requires sufficient energy to overcome the attractive pull. A certain amount of energy is required for an electron to pull it out from the surface of the metal. That minimum energy required is known as the work function of the metal.