Frequency of a photon having energy 66 eV is,
A. \[8 \times {10^{ - 15}}Hz\]
B. \[12 \times {10^{ - 15}}Hz\]
C. \[16 \times {10^{15}}Hz\]
D. None of these
Answer
289.2k+ views
Hint:The frequency of the photon is given. So, to find the energy carried by the photon we use the quantization of energy formula. The energy of the photon is proportional to the frequency of the photon.
Formula used:
\[E = h\nu \],
here $h$ is the Plank’s constant and E is the energy of the photon with frequency equals to \[\nu \].
\[c = \nu \lambda \]
Here c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.
Complete step by step solution:
It is given that the energy of the photon is 66 eV. We need to find the frequency of the photon with the given energy. We need to change the unit of energy from eV to joule. One electron-Volt is the energy used to accelerate an electron in a region of electric potential 1 volt.
So, \[1eV = 1.6 \times {10^{ - 19}}J\]
Hence, the energy of the photon is,
\[E = 66 \times \left( {1.6 \times {{10}^{ - 19}}} \right)J\]
\[\Rightarrow E = 1.06 \times {10^{ - 17}}J\]
Hence, the energy of the given photon is \[1.06 \times {10^{ - 17}}J\].
We got the energy of the photon and now we need to use the formula of energy of the photon to find the frequency of the photon,
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
\[\Rightarrow \nu = \dfrac{E}{h}\]
Putting the values of Plank’s constant and the energy, we get
\[\nu = \dfrac{{\left( {1.06 \times {{10}^{ - 17}}} \right)}}{{\left( {6.626 \times {{10}^{ - 34}}} \right)}}Hz\]
\[\Rightarrow \nu = 1.59 \times {10^{16}}Hz\]
On rounding-off the value to two decimal places, we get the frequency of the photon as,
\[\nu = 1.6 \times {10^{16}}Hz\]
\[\Rightarrow \nu = \left( {1.6 \times 10} \right) \times {10^{15}}Hz\]
\[\therefore \nu = 16 \times {10^{15}}Hz\]
Hence, the frequency of the given photon is \[16 \times {10^{15}}Hz\].
Therefore, the correct option is C.
Note: The given unit of the energy is electron-volt and the frequency is given in Hertz. So, we must change the given energy unit to Joules from electron volt. Then apply the energy formula.
Formula used:
\[E = h\nu \],
here $h$ is the Plank’s constant and E is the energy of the photon with frequency equals to \[\nu \].
\[c = \nu \lambda \]
Here c is the speed of light, \[\nu \] is the frequency of the photon and \[\lambda \] is the wavelength of the light wave.
Complete step by step solution:
It is given that the energy of the photon is 66 eV. We need to find the frequency of the photon with the given energy. We need to change the unit of energy from eV to joule. One electron-Volt is the energy used to accelerate an electron in a region of electric potential 1 volt.
So, \[1eV = 1.6 \times {10^{ - 19}}J\]
Hence, the energy of the photon is,
\[E = 66 \times \left( {1.6 \times {{10}^{ - 19}}} \right)J\]
\[\Rightarrow E = 1.06 \times {10^{ - 17}}J\]
Hence, the energy of the given photon is \[1.06 \times {10^{ - 17}}J\].
We got the energy of the photon and now we need to use the formula of energy of the photon to find the frequency of the photon,
Using the expression for the energy of the photon, we find,
\[E = h\nu \]
\[\Rightarrow \nu = \dfrac{E}{h}\]
Putting the values of Plank’s constant and the energy, we get
\[\nu = \dfrac{{\left( {1.06 \times {{10}^{ - 17}}} \right)}}{{\left( {6.626 \times {{10}^{ - 34}}} \right)}}Hz\]
\[\Rightarrow \nu = 1.59 \times {10^{16}}Hz\]
On rounding-off the value to two decimal places, we get the frequency of the photon as,
\[\nu = 1.6 \times {10^{16}}Hz\]
\[\Rightarrow \nu = \left( {1.6 \times 10} \right) \times {10^{15}}Hz\]
\[\therefore \nu = 16 \times {10^{15}}Hz\]
Hence, the frequency of the given photon is \[16 \times {10^{15}}Hz\].
Therefore, the correct option is C.
Note: The given unit of the energy is electron-volt and the frequency is given in Hertz. So, we must change the given energy unit to Joules from electron volt. Then apply the energy formula.
Recently Updated Pages
JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 2) Chemistry Question Paper with Answer Key

JEE Main Mock Test 2025-26: Oscillations and Waves Free Practice

JEE Main Mock Test 2025-26: Optics Chapter Practice Online

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

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

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

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

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2026-27 PDF Download (Login Required)

Hybridisation in Chemistry – Concept, Types & Applications

