
The photoelectric threshold wavelength for potassium (work function being 2 eV) is
(A) 310 nm
(B) 620 nm
(C) 6200 nm
(D) 3100 nm
Answer
146.7k+ views
Hint The threshold wavelength is the minimum wavelength of the light beam which is incident on the material required for emission of electrons. It is related to the work function with Planck’s constant and speed of light. Substitute the data in the expression to obtain the value of$\lambda $.
Complete step-by-step solution
The minimum amount of energy / work needed to remove an electron from the surface of the metal is called work function. It is denoted by φ. For the emission of electrons from the metal surface minimum frequency of the light used is required at which the emission takes place is called threshold frequency and has its corresponding threshold wavelength.
The work function is given by the formula,
$\varphi = \dfrac{{hc}}{\lambda }$
Where,
h is the Planck’s constant
c is the speed of light
$\lambda $ is the threshold wavelength.
The given data:
$
h = 6.62 \times {10^{ - 34}} \\
c = 3 \times {10^8} \\
\varphi = 2eV = 3.2 \times {10^{ - 19}}J \\
$
On substituting the known data in the above formula, we get
$
\lambda = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.2 \times {{10}^{ - 19}}}} \\
\lambda = 620 \times {10^{ - 9}}m \\
\lambda = 620nm \\
$
The threshold wavelength for potassium is 620 nm. The correct option is B.
Note The work function is given in electron volt (eV) so it has to be converted into joules (J).
$1eV = 1.6 \times {10^{ - 19}}J$
The final answer needs to be written in nanometer nm which is equal to 10-9m
Complete step-by-step solution
The minimum amount of energy / work needed to remove an electron from the surface of the metal is called work function. It is denoted by φ. For the emission of electrons from the metal surface minimum frequency of the light used is required at which the emission takes place is called threshold frequency and has its corresponding threshold wavelength.
The work function is given by the formula,
$\varphi = \dfrac{{hc}}{\lambda }$
Where,
h is the Planck’s constant
c is the speed of light
$\lambda $ is the threshold wavelength.
The given data:
$
h = 6.62 \times {10^{ - 34}} \\
c = 3 \times {10^8} \\
\varphi = 2eV = 3.2 \times {10^{ - 19}}J \\
$
On substituting the known data in the above formula, we get
$
\lambda = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3.2 \times {{10}^{ - 19}}}} \\
\lambda = 620 \times {10^{ - 9}}m \\
\lambda = 620nm \\
$
The threshold wavelength for potassium is 620 nm. The correct option is B.
Note The work function is given in electron volt (eV) so it has to be converted into joules (J).
$1eV = 1.6 \times {10^{ - 19}}J$
The final answer needs to be written in nanometer nm which is equal to 10-9m
Recently Updated Pages
How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

Concept of CP and CV of Gas - Important Concepts and Tips for JEE

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

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