Question

Two identical photocathodes receive light of frequencies ${{f}_{1}}$ and ${{f}_{2}}$. If the velocities of the photoelectrons (of mass m) coming out are ${{v}_{1}}$ and ${{v}_{2}}$, respectively. Then
A. ${{v}_{1}}-{{v}_{2}}={{\left[ \dfrac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right) \right]}^{\dfrac{1}{2}}}$
B. $v_{1}^{2}-v_{2}^{2}=\dfrac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right)$
C. ${{v}_{1}}+{{v}_{2}}={{\left[ \dfrac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right) \right]}^{\dfrac{1}{2}}}$
D. $v_{1}^{2}+v_{2}^{2}=\dfrac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right)$

Answer 1

To solve this question we must know about the formula for the kinetic energy of the photoelectrons when they are emitted from the metal surface. We also need the formula for the kinetic energy of a particle from the mechanics point of view.

Formula used:
$K=hf-\phi$
$K=\dfrac{1}{2}m{{v}^{2}}$

Complete step by step answer:
Let us first understand what photoelectrons are. When light or any other electromagnetic wave falls on the surface of a metal, electrons are emitted from the surface of the metal. These emitted electrons are called photoelectrons.Most of the time, this experiment is done in a photoelectric cell and the metal is called photocathode.

When light is incident on the metal, the free moving electrons near the surface absorb the energy of the light. If the energy is enough is enough to make the interatomic force of attraction, then the electrons are ejected from the metal. The emitted electrons possess some kinetic energy which is given as $K=hf-\phi$, where h is the Planck’s constant, f is the frequency of the light and $\phi$ is the work of the metal. For identical photocathodes (made up of the same metal), the work function is the same.

Now, let the kinetic energy of the photoelectrons for the frequency ${{f}_{1}}$ be ${{K}_{1}}$
And let the kinetic energy of the photoelectrons for the frequency ${{f}_{2}}$ be ${{K}_{2}}$.
$\Rightarrow {{K}_{1}}=h{{f}_{1}}-\phi$ …. (i)
And
${{K}_{2}}=h{{f}_{2}}-\phi$ ….. (ii)
Now, subtract (ii) from (i).
$\Rightarrow {{K}_{1}}-{{K}_{2}}=h{{f}_{1}}-\phi -(h{{f}_{2}}-\phi )=h\left( {{f}_{1}}-{{f}_{2}} \right)$ …. (iii)

We know also that the kinetic energy of a particle is given as $K=\dfrac{1}{2}m{{v}^{2}}$,
where m is its mass (in this case its electron’s mass) and v is its speed.
Therefore,
$\Rightarrow {{K}_{1}}=\dfrac{1}{2}mv_{1}^{2}$ …. (iv)
And
${{K}_{2}}=\dfrac{1}{2}mv_{2}^{2}$ ….. (v)
Now, subtract (v) from (iv).
$\Rightarrow {{K}_{1}}-{{K}_{2}}=\dfrac{1}{2}mv_{1}^{2}-\dfrac{1}{2}mv_{2}^{2}=\dfrac{1}{2}m\left( v_{1}^{2}-v_{2}^{2} \right)$ …. (vi).

From the equation (iii) and (vi), we get that
$\Rightarrow \dfrac{1}{2}m\left( v_{1}^{2}-v_{2}^{2} \right)=h\left( {{f}_{1}}-{{f}_{2}} \right)$
$\therefore \left( v_{1}^{2}-v_{2}^{2} \right)=\dfrac{2h}{m}\left( {{f}_{1}}-{{f}_{2}} \right)$.

Hence, the correct option is B.

Note: You may sometimes see that the kinetic energy of the photoelectrons that we discussed above is written as maximum kinetic energy of a photoelectron. This is because all the photoelectrons have different energies. This means that, after the electron is ejected from the metal, not all the remaining energy is converted in kinetic energy. This happens due the various interactions within the metal like collisions with other electrons decrease their kinetic energy.