
The work function of a photoelectric material is 4 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is \[2.5{\text{ }}V.\]
Answer
124.2k+ views
Hint: The threshold wavelength corresponds to the work function of a photoelectric material and is inversely proportional to it. The stopping corresponds to the potential required to stop electrons ejected from the photoelectric material.
Formula used: In this solution, we will use the following formula:
$\phi = \dfrac{{hc}}{{{\lambda _{thresh}}}}$ where $h$ is the Planck’s constant, $c$ is the speed of light, and ${\lambda _{thresh}}$ is the wavelength of the photon
Complete step by step answer:
We’ve been given the work function of a photoelectric material as 4 eV. The threshold wavelength of the material corresponds to the energy that the incoming photon must have to cause the photoelectric effect. Since the incoming photon must have energy equal to the work function of the material, we can calculate the threshold wavelength as
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 4\, \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 3.1 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 310\,nm\]
Hence the threshold wavelength of the material is 310 nm. The incoming photon must have a wavelength less than or equal to this value.
b) Now we know that the stopping potential is \[2.5{\text{ }}V.\] and we want to find the corresponding threshold wavelength. So again, using the formula
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 2.5 \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 1.91 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 190\,nm\]
Note: The threshold wavelength corresponds to the work function of the material and if a photon corresponding to the threshold wavelength is incident on the material, the ejected electron will have no kinetic energy and will eventually recombine with the metal. To have a non-zero ejected electron velocity, the wavelength of the photon must be less than the threshold wavelength.
Formula used: In this solution, we will use the following formula:
$\phi = \dfrac{{hc}}{{{\lambda _{thresh}}}}$ where $h$ is the Planck’s constant, $c$ is the speed of light, and ${\lambda _{thresh}}$ is the wavelength of the photon
Complete step by step answer:
We’ve been given the work function of a photoelectric material as 4 eV. The threshold wavelength of the material corresponds to the energy that the incoming photon must have to cause the photoelectric effect. Since the incoming photon must have energy equal to the work function of the material, we can calculate the threshold wavelength as
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 4\, \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 3.1 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 310\,nm\]
Hence the threshold wavelength of the material is 310 nm. The incoming photon must have a wavelength less than or equal to this value.
b) Now we know that the stopping potential is \[2.5{\text{ }}V.\] and we want to find the corresponding threshold wavelength. So again, using the formula
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 2.5 \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 1.91 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 190\,nm\]
Note: The threshold wavelength corresponds to the work function of the material and if a photon corresponding to the threshold wavelength is incident on the material, the ejected electron will have no kinetic energy and will eventually recombine with the metal. To have a non-zero ejected electron velocity, the wavelength of the photon must be less than the threshold wavelength.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

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

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

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

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

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

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

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Physics Average Value and RMS Value JEE Main 2025

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