Two streams of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of the ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x :\[3\]. The value of x is ________.
Answer
301.8k+ views
Hint: When a radiation of sufficient energy greater than the work function of the metal is incident on a metal surface, the electrons inside the atoms of metal surface gain energy and eject out of the surface. This is the Einstein theory of radiation.
By planck's law, \[E{\text{ }} = {\text{ }}h\nu \]where 'h' is known as Planck's constant having value \[6.63{\left( {10} \right)^{34}}\]J s, and 'ν' is the frequency of radiation.
According to Einstein photoelectric equation, kinetic energy.\[ = {\text{ }}h\nu {\text{ }}-{\text{ }}\Phi \] where \[\Phi \]is known as work function.
Formula used:
Photo electric work function:
\[E = {E_0} - {\text{ }}\Phi \]
Complete answer:
Combining the two laws, one given by plank and other by Einstein
\[E = {E_0} - {\text{ }}\Phi \]
It is given that, \[E_0\] for the first photon:
\[ = E{0_1} = {\text{ }}2{\text{ }}\Phi \]
(as photon possesses energy equal to twice the work function of metal).
\[E_0\] for second photon
\[ = E{0_2} = {\text{ }}10{\text{ }}\Phi \]
(as second photon possesses energy equal to ten times the work function of metal)
Thus,
\[{E_1} = 2{\text{ }}\Phi {\text{ }}-{\text{ }}\Phi = \Phi \]
\[{E_2} = 10{\text{ }}\Phi {\text{ }}-{\text{ }}\Phi = 9\Phi \]
Kinetic energy also is dependent on velocity and mass by the relation
\[E = \dfrac{1}{2}m{v^2}\]
Where v is the velocity of a photon. Mass is a constant value hence gets cancelled when the ratio of first photon to the second photon is taken.
Therefore,
\[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{V_1}^2}}{{{V_2}^2}}\]
\[\dfrac{1}{9} = \dfrac{{{V_1}^2}}{{{V_2}^2}}\]
The ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : 3.
Thus,
\[\dfrac{1}{3} = \dfrac{{{V_1}}}{{{V_2}^{}}}\]
\[\dfrac{1}{3} = \dfrac{x}{3}\]
$x = 3$
Thus, the value of x is $3$.
Note: Minimum energy required for emission of electrons from the surface of metal is called the threshold frequency. The speed and number of the emitted electrons will depend on two factors, namely the colour and intensity of the incident radiation.
By planck's law, \[E{\text{ }} = {\text{ }}h\nu \]where 'h' is known as Planck's constant having value \[6.63{\left( {10} \right)^{34}}\]J s, and 'ν' is the frequency of radiation.
According to Einstein photoelectric equation, kinetic energy.\[ = {\text{ }}h\nu {\text{ }}-{\text{ }}\Phi \] where \[\Phi \]is known as work function.
Formula used:
Photo electric work function:
\[E = {E_0} - {\text{ }}\Phi \]
Complete answer:
Combining the two laws, one given by plank and other by Einstein
\[E = {E_0} - {\text{ }}\Phi \]
It is given that, \[E_0\] for the first photon:
\[ = E{0_1} = {\text{ }}2{\text{ }}\Phi \]
(as photon possesses energy equal to twice the work function of metal).
\[E_0\] for second photon
\[ = E{0_2} = {\text{ }}10{\text{ }}\Phi \]
(as second photon possesses energy equal to ten times the work function of metal)
Thus,
\[{E_1} = 2{\text{ }}\Phi {\text{ }}-{\text{ }}\Phi = \Phi \]
\[{E_2} = 10{\text{ }}\Phi {\text{ }}-{\text{ }}\Phi = 9\Phi \]
Kinetic energy also is dependent on velocity and mass by the relation
\[E = \dfrac{1}{2}m{v^2}\]
Where v is the velocity of a photon. Mass is a constant value hence gets cancelled when the ratio of first photon to the second photon is taken.
Therefore,
\[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{V_1}^2}}{{{V_2}^2}}\]
\[\dfrac{1}{9} = \dfrac{{{V_1}^2}}{{{V_2}^2}}\]
The ratio of maximum velocities of the photoelectrons emitted in the two respective cases is x : 3.
Thus,
\[\dfrac{1}{3} = \dfrac{{{V_1}}}{{{V_2}^{}}}\]
\[\dfrac{1}{3} = \dfrac{x}{3}\]
$x = 3$
Thus, the value of x is $3$.
Note: Minimum energy required for emission of electrons from the surface of metal is called the threshold frequency. The speed and number of the emitted electrons will depend on two factors, namely the colour and intensity of the incident radiation.
Recently Updated Pages
Page Not Found - NCERT Solutions, CBSE, JEE, NEET 2026-27

Single Slit Diffraction - Young’s Single Slit Experiment

Carbohydrates Class 12 Important Questions JEE Advanced Chemistry [PDF]

Crack JEE Advanced 2026 with Vedantu's Live Classes

JEE Advanced 2027 Revision Notes for Amino Acids and Peptides - Free PDF Download

JEE Advanced 2027 Revision Notes for Chemical Equilibrium - Free PDF Download

Trending doubts
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

IIT CSE Cutoff 2026: Opening and Closing Ranks for B.Tech. Computer Science Admissions at All IITs

IIT Fees Structure for BTech Courses 2026: Complete Details

Top IIT Colleges in India 2026: List, Rankings, Fees, Placements & Admission

JEE Advanced Previous Year Question Papers with Solutions PDF (2012–2026)

Other Pages
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

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

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

