
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
162.9k+ views
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.
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
