Question

A photon of energy $8.6eV$ is incident on a metal surface of threshold frequency \[1.6 \times {10^{15}}Hz\] . The maximum kinetic energy of the photons emitted (in$eV$) is nearly:
(A) $1.6$
(B) $6$
(C) $2$
(D) $1.2$

Answer 1

Hint: We will use the concept that the total energy incident on a surface equals the threshold frequency plus the maximum kinetic energy of the emitted photons.
Formulae Used \[h\nu = h{\nu _0} + K.{E_{max}}\]
Where, $h$ is the Planck's constant having a value of \[4.14 \times {10^{ - 15}}eVH{z^{ - 1}}\].

Step By Step Solution
Here,
Given,
\[h\nu = 8.6eV\]
\[{\nu _0} = 1.6 \times {10^{15}}Hz\]
Thus,
\[h{\nu _0} = 6.62eV\]
Hence,
\[K.E{._{max}} = 8.60 - 6.62 = 1.98eV\]

Additional InformationWhen a radiation of certain frequency strikes the surface of a metal, particles are emitted by the surface. These particles were called photons.
Firstly it was thought that it emits some electrons and protons under some circumstances but after experimental procedure, it was found that this was not the case.
Then, Einstein came as a savior and saved the day by giving some theories on this phenomenon which tallied with the experimental data. The main theory was that electrons are emitted in small packets of energy called photons.
 This effect was named in a very intuitive manner.
Firstly, a light radiation is incident on the surface, thus the name photo. Thereafter electrons are emitted, thus the name electric. Hence, the total name turned out at photoelectric effect.

NoteThe value we got signifies the maximum kinetic energy a photon could attain in this circumstance. Also, through the formula, we can say that after the energy is incident on the surface, the photons first break the barrier of the threshold frequency and then the left out energy is turned into the kinetic energy of the photons.