
Ultraviolet radiation of 6.2eV falls on an aluminium surface (work function 4.2eV). Then find the kinetic energy (in joule) of the fastest electron emitted.
A. \[3.2 \times {10^{ - 21}}J\]
B. \[1.6 \times {10^{ - 17}}J\]
C. \[3.2 \times {10^{ - 19}}J\]
D. \[3.2 \times {10^{ - 15}}J\]
Answer
221.1k+ views
Hint: When light is incident on a metal surface, electrons are emitted known as photoelectrons. This phenomenon is called the photoelectric effect. Work function is the minimum energy required to remove one electron from a metal surface. Work function can also be said as the characteristics of the surface and not the complete metal. As we know that the energy of the incident light or photons should be more than the work function of the metal for the photoemission to take place.
Formula Used:
To find the kinetic energy of the emitted electron we have,
\[K.E = E - W\]
Where, W is the work function and E is the energy of a photon.
Complete step by step solution:
As we know that the kinetic energy of the emitted electron is,
\[K.E = E - W\]
Here, the energy of radiation is 6.2eV and the work function is 4.2eV. Since both the values are in eV, we need to convert it into joules by multiplying it by \[1.6 \times {10^{ - 19}}J\]. Then,
\[E = 6.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow E = 9.92 \times {10^{ - 19}}J\]and
\[\Rightarrow W = 4.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow W = 6.72 \times {10^{ - 19}}J\]
Now, substitute these value sin above equation, we get,
\[K.E = 9.92 \times {10^{ - 19}} - 6.72 \times {10^{ - 19}}\]
\[\therefore K.E = 3.2 \times {10^{ - 19}}J\]
Therefore, the kinetic energy of the fastest electron emitted is \[3.2 \times {10^{ - 19}}J\].
Hence, Option C is the correct answer.
Note:Remember that, if the energy is given in eV, then we need to convert the given value of energy from eV to joules by multiplying the charge of an electron while solving this problem. Moreover, Metals contain many free electrons that are loosely bound to the atom. The typical value of work function varies from 2ev to 6eV.
Formula Used:
To find the kinetic energy of the emitted electron we have,
\[K.E = E - W\]
Where, W is the work function and E is the energy of a photon.
Complete step by step solution:
As we know that the kinetic energy of the emitted electron is,
\[K.E = E - W\]
Here, the energy of radiation is 6.2eV and the work function is 4.2eV. Since both the values are in eV, we need to convert it into joules by multiplying it by \[1.6 \times {10^{ - 19}}J\]. Then,
\[E = 6.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow E = 9.92 \times {10^{ - 19}}J\]and
\[\Rightarrow W = 4.2 \times 1.6 \times {10^{ - 19}}J\]
\[\Rightarrow W = 6.72 \times {10^{ - 19}}J\]
Now, substitute these value sin above equation, we get,
\[K.E = 9.92 \times {10^{ - 19}} - 6.72 \times {10^{ - 19}}\]
\[\therefore K.E = 3.2 \times {10^{ - 19}}J\]
Therefore, the kinetic energy of the fastest electron emitted is \[3.2 \times {10^{ - 19}}J\].
Hence, Option C is the correct answer.
Note:Remember that, if the energy is given in eV, then we need to convert the given value of energy from eV to joules by multiplying the charge of an electron while solving this problem. Moreover, Metals contain many free electrons that are loosely bound to the atom. The typical value of work function varies from 2ev to 6eV.
Recently Updated Pages
JEE Main 2022 (July 26th Shift 1) Physics Question Paper with Answer Key

JEE Main 2025-26: Dual Nature of Matter and Radiation Mock Test

JEE Main 2025-26 Electromagnetic Waves Mock Test with Solutions

JEE Main 2022 (June 26th Shift 2) Chemistry Question Paper with Answer Key

Apparent Frequency Explained: Formula, Uses & Examples

JEE Main 2023 (January 30th Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Understanding Electromagnetic Waves and Their Importance

