Question

Work function of a metal is \[2.1eV\]. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surface?
A. \[4000\mathop A\limits^0 ,{\rm{ 7500}}\mathop A\limits^0 \]
B. \[5500\mathop A\limits^0 ,{\rm{ 6000}}\mathop A\limits^0 \]
C. \[4000\mathop A\limits^0 ,{\rm{ 6000}}\mathop A\limits^0 \]
D. None of these

Answer 1

Hint: Work function is defined as the minimum energy which is required by an electron to escape from the metal surface. On the other hand, we say that the work function is the energy that is needed to eject electrons from the metal surface. So, the maximum energy (E) of the photoelectron can be equal to the energy of the incident photon (h) minus the work function (\[\phi \]), because an electron has to do some work to escape the potential from the metallic surface.

Formula used:
Work function is given as:
\[\phi = h{\upsilon _0} = \dfrac{{hc}}{{{\lambda _0}}}\]
Where, h is Planck's constant, \[{\upsilon _0}\] is the threshold frequency, c is the speed of light and \[{\lambda _0}\] is the wavelength.

Complete step by step solution:
Given work function of a metal = \[2.1eV\]
As we know that work function,
\[\phi = \dfrac{{hc}}{{{\lambda _0}}}\]
\[\Rightarrow {\lambda _0} = \dfrac{{hc}}{\phi }\]
Also, we know;
\[E(eV) = \dfrac{{hc}}{\lambda } = \dfrac{{12375}}{{\lambda (\mathop A\limits^0 )}}\]
After substituting the values,
\[{\lambda _0} = \dfrac{{12375}}{{2.1}} = 5892.8\mathop A\limits^0 \]
For the emission of photoelectron from its surface, the condition follows is
\[\lambda < {\lambda _0}\]
In option A, the first value follows but not the second one so is not the correct option. Similarly, options B and C both are not correct.

Hence option D is the correct answer.

Note: As we know the work function is inversely related to the wavelength. So, if the work function increases then wavelength decreases and if the work function decreases then the wavelength increases. Also, the photons are defined as packets of energy and the energy required to knock out electrons is to be provided all at once. In order to emit an electron from the surface of the metal the energy of the electron must be greater than the work function.