
In photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelength of incident light is 260nm, the maximum kinetic energy electrons will be:
Given \[E\left( {{\text{in eV}}} \right) = \dfrac{{1237}}{{\lambda ({\text{in nm)}}}}\]
(A) \[1.5eV\]
(B) \[4.5eV\]
(C) \[15.1eV\]
(D) \[3.0eV\]
Answer
145.8k+ views
Hint: The maximum kinetic energy of ejected electrons in a photoelectric experiment can be given by the energy of the radiation minus the work function of the metal. The work function is an energy threshold which is the minimum energy required to eject the electron without giving it a kinetic energy.
Formula used: In this solution we will be using the following formulae;
\[K{E_{\max }} = E - {E_0}\] where \[K{E_{\max }}\] is the maximum kinetic energy of an electron, \[E\] is the energy of the incident radiation, and \[{E_0}\] is the work function (threshold energy) of the metal.
Complete Step-by-Step Solution:
We are told that a light of wavelength of 260 nm used as an incident light in a photoelectric experiment. Whatever the metal that was used, we are informed that the light must have a threshold wavelength of 380 nm (maximum wavelength which will eject an electron). We are to determine the maximum kinetic energy of the electrons.
We note that the kinetic energy can be given as
\[K{E_{\max }} = E - {E_0}\] where \[K{E_{\max }}\] is the maximum kinetic energy of an electron, \[E\] is the energy of the incident radiation, and \[{E_0}\] is the work function (threshold energy) of the metal. This can be written as
\[K{E_{\max }} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}\] since \[E = \dfrac{{hc}}{\lambda }\] where \[h\] is Planck’s constant and \[c\] is speed of light, \[\lambda \] is the wavelength.
As given, \[E\left( {{\text{in eV}}} \right) = \dfrac{{1237}}{{\lambda ({\text{in nm)}}}}\]
Then,
\[K{E_{\max }} = \dfrac{{1237}}{{260}} - \dfrac{{1237}}{{380}}\]
By computation, we have
\[K{E_{\max }} = 1.5eV\]
Hence, the correct option is A
Note: Although it may be odd or confusing, observe that the minimum energy to eject a photon corresponds to the maximum wavelength required. This is because the wavelength and energy are inversely related.
Formula used: In this solution we will be using the following formulae;
\[K{E_{\max }} = E - {E_0}\] where \[K{E_{\max }}\] is the maximum kinetic energy of an electron, \[E\] is the energy of the incident radiation, and \[{E_0}\] is the work function (threshold energy) of the metal.
Complete Step-by-Step Solution:
We are told that a light of wavelength of 260 nm used as an incident light in a photoelectric experiment. Whatever the metal that was used, we are informed that the light must have a threshold wavelength of 380 nm (maximum wavelength which will eject an electron). We are to determine the maximum kinetic energy of the electrons.
We note that the kinetic energy can be given as
\[K{E_{\max }} = E - {E_0}\] where \[K{E_{\max }}\] is the maximum kinetic energy of an electron, \[E\] is the energy of the incident radiation, and \[{E_0}\] is the work function (threshold energy) of the metal. This can be written as
\[K{E_{\max }} = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}}\] since \[E = \dfrac{{hc}}{\lambda }\] where \[h\] is Planck’s constant and \[c\] is speed of light, \[\lambda \] is the wavelength.
As given, \[E\left( {{\text{in eV}}} \right) = \dfrac{{1237}}{{\lambda ({\text{in nm)}}}}\]
Then,
\[K{E_{\max }} = \dfrac{{1237}}{{260}} - \dfrac{{1237}}{{380}}\]
By computation, we have
\[K{E_{\max }} = 1.5eV\]
Hence, the correct option is A
Note: Although it may be odd or confusing, observe that the minimum energy to eject a photon corresponds to the maximum wavelength required. This is because the wavelength and energy are inversely related.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

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

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

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

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Ideal and Non-Ideal Solutions Raoult's Law - JEE
