
The light rays having photons of energy 1.8 eV are falling on metal surface having a work function 1.2 eV. What is the stopping potential to be applied in order to stop the emitting electrons?
A. 3V
B. 1.2V
C. 0.6V
D. 1.4V
Answer
127.8k+ views
Hint: To find the stopping potential, we can use the work-energy theorem which states that the net work done is equal to the change in kinetic energy. So, the work done by the applied potential is such that the final kinetic energy of the photoelectron becomes zero. That potential is called the stopping potential.
Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
Complete step by step solution:
The kinetic energy of the photo-electron is the remaining energy of the photon transferred to the ejected electron in the form of kinetic energy. So, the kinetic energy of the photoelectron is calculated by finding the difference of total energy of the photon and the work function of the metal.
When the photoelectron stops on application of the potential difference, then the magnitude of the change in kinetic energy is equal to the maximum kinetic energy of the photoelectron. So, the work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }}\]
The energy of the photon is given as 1.8 eV and the work function of the metal surface is 1.2 eV. The maximum kinetic energy of the ejected electron will be,
\[K = 1.8eV - 1.2eV\]
\[\Rightarrow K = 0.6\,eV\]
So, the stopping potential will be,
\[e{V_0} = 0.6\,eV\]
\[\therefore {V_0} = 0.6V\]
Hence, the stopping potential is 0.6V.
Therefore, the correct option is C.
Note: While using the work-energy theorem, we don’t need to put the negative sign in the value of the change in kinetic energy of the photoelectron. So, the stopping potential energy is equal to the kinetic energy of the photoelectron.
Formula used:
\[K = \dfrac{{hc}}{\lambda } - \phi \]
Here K is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[\phi \] is the work function of the metal.
Complete step by step solution:
The kinetic energy of the photo-electron is the remaining energy of the photon transferred to the ejected electron in the form of kinetic energy. So, the kinetic energy of the photoelectron is calculated by finding the difference of total energy of the photon and the work function of the metal.
When the photoelectron stops on application of the potential difference, then the magnitude of the change in kinetic energy is equal to the maximum kinetic energy of the photoelectron. So, the work done by the stopping potential of \[{V_0}\] to stop a moving electron must be equal to the kinetic energy of the electron.
\[e{V_0} = K{E_{\max }}\]
The energy of the photon is given as 1.8 eV and the work function of the metal surface is 1.2 eV. The maximum kinetic energy of the ejected electron will be,
\[K = 1.8eV - 1.2eV\]
\[\Rightarrow K = 0.6\,eV\]
So, the stopping potential will be,
\[e{V_0} = 0.6\,eV\]
\[\therefore {V_0} = 0.6V\]
Hence, the stopping potential is 0.6V.
Therefore, the correct option is C.
Note: While using the work-energy theorem, we don’t need to put the negative sign in the value of the change in kinetic energy of the photoelectron. So, the stopping potential energy is equal to the kinetic energy of the photoelectron.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Difference Between Solute and Solvent: JEE Main 2024

Absolute Pressure Formula - Explanation, and FAQs

Carbon Dioxide Formula - Definition, Uses and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

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

JEE Main Login 2045: Step-by-Step Instructions and Details

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Diffraction of Light - Young’s Single Slit Experiment
