Question

On a photosensitive material, when the frequency of incident radiation is increased by 30% kinetic energy of emitted photoelectrons increases from 0.4eV to 0.9eV. The work function of the surface is:
A. 1eV
B. 1.267eV
C. 1.4eV
D. 1.8eV.

Answer 1

Hint: The photoelectric equation was formulated by Einstein. The entire photon energy is transferred to the electron when the photon falls on the metal surface.

Formula used:
\[h\nu = {\text{ }}{\phi _0} + {K_{\max }}\]
Where, $h$ is Planck’s constant $h = 6.625 \times {10^{ - 34}}Js$, $E = h\nu$ is energy of the photon, $c$ is speed of light, $\nu$ is frequency of incident light, \[{\phi _0}\] work function, \[{{\text{K}}_{{\text{max}}}}\] is maximum kinetic energy of an electron.

Complete answer:
Let us consider,
Case(i): Initial frequency of incident radiation be ${\nu _1}$and initial kinetic energy of emitted photoelectrons \[{K_1} = 0.4eV\]
Apply Einstein’s photoelectric equation,
\[h\nu = {\text{ }}{\phi _0} + {K_{\max }}\]
\[h{\nu _1} = {\text{ }}{\phi _0} + 0.4eV\] ………………. (1)
From case 1 we get the value for the initial frequency of initial radiation.
Now consider,
Case(ii): final frequency of incident radiation be ${\nu _2}$and initial kinetic energy of emitted photoelectrons \[{K_{_2}} = 0.9eV\]
Then, the frequency of incident radiation is increased by 30% .Therefore,
${\nu _2} = {\nu _1} + \dfrac{{30}}{{100}}{\nu _1}$
${\nu _2} = {\nu _1} + 0.3{\nu _1} \\$
On simplification, we get
${\nu _2} = 1.3{\nu _1} \\ $
We shall now apply Einstein’s photoelectric equation,
\[h{\nu _2} = {\phi _0} + {K_2}\]
\[h\left( {1.3{\nu _1}} \right) = {\phi _0} + 0.9eV\]
\[{\text{1}}{\text{.3h}}{\nu _1}{\text{ = }}{\phi _0} + 0.9{\text{eV}}\] ………….. (2)
Now substitute the value of \[h{\nu _1}\]in equation (2),
We get, \[1.3({\phi _0} + 0.4eV) = {\text{ }}{\phi _0} + 0.9eV\]
\[1.3{\phi _0} + 0.52 = {\phi _0} + 0.9\]
On simplification, we get
\[1.3{\phi _0} - {\phi _0} = 0.9 - 0.52\]
\[{\phi _0} = 1.267eV\]

$\therefore$ The Work function of the surface is \[1.267{\text{eV}}\]. Therefore, correct option is (B).

Additional information:
According to Einstein’s photoelectric equation, when a light is an incident on metal, the photons having energy \[{\text{h\nu }}\]collide with electrons at the surface of the metal. During these collisions, the energy of the photon is completely transferred to the electron. If this energy is sufficient, the electrons are ejected out of the metal instantaneously. The minimum energy needed for the electron to come out of the metal surface is called work function. If the energy \[{\text{h\nu }}\] of the incident photon exceeds the work function\[\left( {{\phi _0}} \right)\] ,the electrons are emitted with a maximum kinetic energy.
\[h\nu = {\phi _0} + {K_{\max }}\]
\[{K_{\max }} = h\nu - {\phi _0}\]

Note:
The gain of the kinetic energy of an electron is the difference between the work function of the metal and the energy of the incident photon. The energy required to remove the electron from the material is known as work function.