
Light of wavelength $\lambda $ falls on a metal having work function $\dfrac{{hc}}{{{\lambda _0}}}$. Photoelectric effect will take place only if:
A) $\lambda \geqslant {\lambda _0}$
B) $\lambda \geqslant 2{\lambda _0}$
C) $\lambda \leqslant {\lambda _0}$
D) $\lambda < \dfrac{{{\lambda _0}}}{2}$
Answer
149.1k+ views
Hint: Recall that the photoelectric effect is a phenomenon in which the electrons are ejected from the surface of a metal when the light is made to fall on the surface of the metal. These ejected electrons are known as photoelectrons. The electrons absorb energy and get ejected.
Complete step by step solution:
When light is incident on the surface of metal, the electrons absorb some amount of light. This is because some amount of light is used as energy and the rest of the energy is used to increase kinetic energy of the electrons ejected.
According to Einstein equation of photoelectric effect, the maximum kinetic energy of the electrons can be written as:
$K.E. = h\nu - h{\nu _0}$---(i)
The above equation can also be written as
$h\nu = h{\nu _0} + K.E.$
Given that the wavelength of the light is $ = \lambda $
The work function is $\phi = \dfrac{{hc}}{{{\lambda _0}}}$
The energy of the metal is $E = \dfrac{{hc}}{\lambda }$
The photoelectric effect will take place only if the energy of this effect will be greater than or equal to the work function of the metal.
$\therefore $ As per the condition of photoelectric effect, it can be written that
$ \Rightarrow \dfrac{{hc}}{\lambda } \geqslant \dfrac{{hc}}{{{\lambda _0}}}$
Or it can be written that
$\dfrac{1}{\lambda } \geqslant \dfrac{1}{{{\lambda _0}}}$
Or $\lambda \leqslant {\lambda _0}$
$\therefore $ Photoelectric effect will take place only if $\lambda \leqslant {\lambda _0}$
Option C is the right answer.
Note: It is important to remember that the work function of a metal is defined as the minimum amount of energy that is required to remove an electron from the surface of the metal. The kinetic energy of the incident light varies with the intensity. This means the energy increases if the intensity of light is increased. But the light that has low intensity also ejects electrons. It does so after some time so that the electrons acquire enough energy to get emitted.
Complete step by step solution:
When light is incident on the surface of metal, the electrons absorb some amount of light. This is because some amount of light is used as energy and the rest of the energy is used to increase kinetic energy of the electrons ejected.
According to Einstein equation of photoelectric effect, the maximum kinetic energy of the electrons can be written as:
$K.E. = h\nu - h{\nu _0}$---(i)
The above equation can also be written as
$h\nu = h{\nu _0} + K.E.$
Given that the wavelength of the light is $ = \lambda $
The work function is $\phi = \dfrac{{hc}}{{{\lambda _0}}}$
The energy of the metal is $E = \dfrac{{hc}}{\lambda }$
The photoelectric effect will take place only if the energy of this effect will be greater than or equal to the work function of the metal.
$\therefore $ As per the condition of photoelectric effect, it can be written that
$ \Rightarrow \dfrac{{hc}}{\lambda } \geqslant \dfrac{{hc}}{{{\lambda _0}}}$
Or it can be written that
$\dfrac{1}{\lambda } \geqslant \dfrac{1}{{{\lambda _0}}}$
Or $\lambda \leqslant {\lambda _0}$
$\therefore $ Photoelectric effect will take place only if $\lambda \leqslant {\lambda _0}$
Option C is the right answer.
Note: It is important to remember that the work function of a metal is defined as the minimum amount of energy that is required to remove an electron from the surface of the metal. The kinetic energy of the incident light varies with the intensity. This means the energy increases if the intensity of light is increased. But the light that has low intensity also ejects electrons. It does so after some time so that the electrons acquire enough energy to get emitted.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application

Young's Double Slit Experiment Step by Step Derivation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Uniform Acceleration

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Electrical Field of Charged Spherical Shell - JEE

Charging and Discharging of Capacitor
