Question

Photons of energy 1eV and 2.5eV successively illuminate a metal whose work function is 0.5eV. The ratio of the maximum speed of emitted electrons is…….
A. 1:2
B. 2:1
C. 3:1
D. 1:3

Answer 1

Hint: Define the photoelectric effect. Write Einstein's photoelectric equation. From this relation first, find the ratio of kinetic energies of both the electrons. Then using the formula of kinetic energy in terms of velocity, find out the ratio of emitted electrons.

Complete step-by-step answer:
Photoelectric effect:
When the light (photon) of sufficient energy hits the metal surface, electrons are emitted from the surface of the metal. This phenomenon is called the Photoelectric effect.
In this phenomenon, when light (photon) of sufficient energy hits a metal surface, some part of the energy is used to knock out the electron from its orbit and rest of the energy is gained by the electron which appears in the form of the kinetic energy.
The energy required to knock out the electron from its outermost orbit from the metal surface is known as the work function of that metal.
Thus, we have
Energy of photon = work function + kinetic energy
Let $h\nu$ be the energy of the photon and $\varphi$ be the work function of the metal. Then we have,
 $h\nu =\varphi +K.E.$
Thus,
 $K.E.=h\nu -\varphi$
Given, $\varphi =0.5eV$
For photon 1,
 $h\nu =1eV$
Therefore,
 $\begin{array}{rl}& K.E{.}_1=1eV-0.5eV\\ & K.E{.}_1=0.5eV\end{array}$

For photon 2,
 $h\nu =2.5eV$
Therefore,
$\begin{array}{rl}& K.E{.}_1=2.5eV-0.5eV\\ & K.E{.}_1=2eV\end{array}$

Taking the ratio of these two kinetic energies,
 ${\displaystyle \frac{K.E{.}_1}{K.E{.}_1}}={\displaystyle \frac{0.5}{2}}={\displaystyle \frac{1}{4}}$
Let $v_1$ and $v_2$ be the velocities of the electron when photon 1 and photon 2 are incident on the metal surface respectively.
Kinetic energy of the electron is given as
 $K.E.={\displaystyle \frac{1}{2}}m{v}^2$
Thus,
$${\displaystyle \frac{{\displaystyle \frac{1}{2}}m{v_1}^2}{{\displaystyle \frac{1}{2}}m{v_2}^2}}={\displaystyle \frac{1}{4}}$$
$$\begin{array}{rl}& {\displaystyle \frac{{v_1}^2}{{v_2}^2}}={\displaystyle \frac{1}{4}}\\ & \therefore {\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{1}{2}}\end{array}$$
Thus, the ratio of the maximum speed of the emitted electron is 1:2.
Answer- A. 1:2

Note: Note that if the energy of the photon is less than the work function of the metal then electrons are not emitted from the metal surface. The electrons emitted from the metal surface are called photoelectrons and these electrons are the valence electrons i.e. electrons in outermost orbit. The kinetic energy of the emitted photoelectron depends on the wavelength or frequency of the light, not on its intensity whereas the number of photoelectrons emitted depends upon the intensity of light.