
In photo-electric effect, when photons of energy $'h\nu '$ are incident on a metal surface, electrons are emitted with some kinetic energy. It is possible to say that
(A) All ejected electrons have the same K.E. equal to \[h{\nu _0}\]
(B) The ejected electrons have a distribution of K.E. from zero to $h\nu - h{\nu _0}$
(C) The most energetic electrons have K.E. equal to $h\nu $
(D) All ejected electrons have K.E. equal to $h\nu $
Answer
240.6k+ views
Hint: To solve this question, we need to use Einstein's photoelectric equation. From that equation we can find out the value of the kinetic energy of the most energetic electrons.
Complete step-by-step solution:
We know that when radiation is incident on a metal surface, electrons are not ejected until the photons present in the radiation have the energy equal to or greater than the work function ${\varphi _0}$ of the metal. If the threshold frequency of the metal surface is equal to ${\nu _0}$, then the work function of the metal is given by
${\varphi _0} = h{\nu _0}$
So the photons must carry energy greater than or equal to \[h{\nu _0}\] . If the photons in the radiation have a frequency of $\nu $, then the energy of the incident radiation becomes equal to $h\nu $. Now, from the Einstein’s photoelectric equation, the kinetic energy of the ejected photoelectrons is given by
$K = h\nu - h{\nu _0}$
But we must note that this value of the kinetic energy is possessed by the electrons present at the outermost surface of the metal. This is due to the fact that the electrons embedded inside the metal have more potential energy. So they require more energy to get free. So the outermost electrons are the most energetic electrons and they have the kinetic energy equal to $h\nu - h{\nu _0}$.
As we go deeper inside the surface of the metal, the kinetic energy of the electrons decreases rapidly to zero.
Thus, the ejected electrons have a distribution of K.E. from zero to $h\nu - h{\nu _0}$.
Hence, the correct answer is option B.
Note: The photoelectric effect, which is discussed in this question, has many applications in different fields. The examples include the Photostats, which are used to generate the copies of a printed paper. Also, it is used in the photovoltaic cells, which are used in the solar panels for converting the solar energy of the sun into the electricity.
Complete step-by-step solution:
We know that when radiation is incident on a metal surface, electrons are not ejected until the photons present in the radiation have the energy equal to or greater than the work function ${\varphi _0}$ of the metal. If the threshold frequency of the metal surface is equal to ${\nu _0}$, then the work function of the metal is given by
${\varphi _0} = h{\nu _0}$
So the photons must carry energy greater than or equal to \[h{\nu _0}\] . If the photons in the radiation have a frequency of $\nu $, then the energy of the incident radiation becomes equal to $h\nu $. Now, from the Einstein’s photoelectric equation, the kinetic energy of the ejected photoelectrons is given by
$K = h\nu - h{\nu _0}$
But we must note that this value of the kinetic energy is possessed by the electrons present at the outermost surface of the metal. This is due to the fact that the electrons embedded inside the metal have more potential energy. So they require more energy to get free. So the outermost electrons are the most energetic electrons and they have the kinetic energy equal to $h\nu - h{\nu _0}$.
As we go deeper inside the surface of the metal, the kinetic energy of the electrons decreases rapidly to zero.
Thus, the ejected electrons have a distribution of K.E. from zero to $h\nu - h{\nu _0}$.
Hence, the correct answer is option B.
Note: The photoelectric effect, which is discussed in this question, has many applications in different fields. The examples include the Photostats, which are used to generate the copies of a printed paper. Also, it is used in the photovoltaic cells, which are used in the solar panels for converting the solar energy of the sun into the electricity.
Recently Updated Pages
JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

JEE Main 2025-26 Atoms and Nuclei Mock Test – Free Practice Online

JEE Main 2025-26: Magnetic Effects of Current & Magnetism Mock Test

JEE Main Mock Test 2025: Properties of Solids and Liquids

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Clemmensen and Wolff Kishner Reductions Explained for JEE & NEET

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding the Angle of Deviation in a Prism

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Class 12 Physics Question Paper Set 3 (55/2/3) 2025: PDF, Answer Key & Solutions

CBSE Class 12 Physics Question Paper Set 3 (55/1/3) 2025 – PDF, Solutions & Analysis

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

