
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
232.8k+ 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
JEE Main 2026 Session 2 Registration Open, Exam Dates, Syllabus & Eligibility

JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

Trending doubts
JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Why does capacitor block DC and allow AC class 12 physics JEE_Main

Understanding Average and RMS Value in Electrical Circuits

Understanding Collisions: Types and Examples for Students

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding Atomic Structure for Beginners

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

CBSE Class 12 Physics Set 2 (55/2/2) 2025 Question Paper & Solutions

Inductive Effect and Its Role in Acidic Strength

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Units and Measurements Mock Test for JEE Main 2025-26 Preparation

Chemistry Question Papers for JEE Main, NEET & Boards (PDFs)

