Question

The work function of a photoelectric material is $3.3{\text{eV}}$ . Its threshold frequency will be:
A.$8 \times {10^{14}}{\text{Hz}}$
B.$8 \times {10^{10}}{\text{Hz}}$
C.$5 \times {10^{33}}{\text{Hz}}$
D.$4 \times {10^{11}}{\text{Hz}}$

Answer 1

To answer this question, you must recall the photoelectric effect. For each metal, there is a characteristic minimum frequency known as the threshold frequency below which the photoelectric effect does not occur.
Formula used:
${{\text{W}}_0} = {\text{h}}{\nu _0}$
Where, ${{\text{W}}_0}$ is the work function of the metal.
${\text{h}}$ is the Planck’s constant.
And ${\nu _0}$ is the threshold frequency.

Complete step by step answer:
In photoelectric effect, when electromagnetic radiation, such as light, hits a material, electrons are emitted. Electrons emitted in this process are known as photoelectrons. This effect is studied in solid state, condensed matter physics, and quantum chemistry to figure out conclusions about the properties of atoms, molecules and solids. The effect is commonly used in electronic devices specialized for light detection and precisely timed electron emission.
When a quantum of light strikes a metal surface, it transfers its energy to the electrons in the metal. In order for the electron to escape from the surface of the metal, it must overcome the attractive force of the positive ions in metal. As a result, a part of the photon’s energy is absorbed by the metal surface releasing the electron. This absorbed energy is known as the work function of a metal. The remaining part of the energy of the photon is transferred into the kinetic energy of the electron emitted.
$1{\text{eV}} = 1.6 \times {10^{19}}{\text{J}}$
We know that the work function is given by, ${W_0} = {\text{h}}{\nu _0}$ .
Thus, the threshold frequency is ${\nu _0} = \dfrac{{{W_0}}}{{\text{h}}}$
Substituting the values, we get.
${\nu _0} = \dfrac{{3.3 \times 1.6 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}}$
$\Rightarrow {\nu _0} = 8.434 \times {10^{14}}$
Approximating the value:
$\therefore {\nu _0} \approx 8 \times {10^{14}}{\text{Hz}}$

Hence option A is correct.

Note:
Planck's quantum theory of light accounts for the photoelectric effect. He proposed a hypothesis that the radiant energy such as heat or light, is not emitted continuously but discontinuously in the form of small packets of energy called quanta.