
Line spectrum is obtained from the:
A) Sun
B) Filament of the bulb
C) Mercury lamp
D) Burning coal
Answer
138.9k+ views
Hint: The solution of the given question is totally based on the understanding of line spectrum. First of all we need to know about the spectrum and then about the line spectrum. Also, we need to gather information about the production of the line spectrum and the sources of its production. Then finally, we can conclude with its solution.
Complete answer:
First of all, let us understand the spectrum. We can define spectrum as a beam of light that can be separated in the form of light waves or radio waves. The common example of visible spectrum is the rainbow in which we are able to see different colours separated by a thin boundary.
Now, let us find the line spectrum. We can define a line spectrum as a dark or bright line which results either from the emission or absorption of light in a narrow frequency range.
Now let us discuss how a line spectrum is produced. When there is an interaction between atoms, molecules and a photon of specific frequency, it creates a line spectrum in which the photon is absorbed by the electron and is detected by an observer as a separate line.
Line spectrum can be emitted from arc lamps or other systems in the form of electrical discharge in a sealed gas tube. Neon lamps, hollow cathode lamps, electrode less lamps are some of the examples through which line spectrum can be emitted.
As it can be seen in step two that the source of emission of line spectrum is a sealed gas tube, so from the given options, mercury lamp satisfies this criteria.
Hence, option (C), i.e. Mercury lamp is the correct choice for the given question.
Note: We should not get confused with the filament of the bulb, as it is just a part of the bulb. If in the option it would have been given only a bulb, in that case it can be considered as a source of line spectrum emission. The spectrum of different wavelengths are classified as different bands and known as electromagnetic spectrum range.
Complete answer:
First of all, let us understand the spectrum. We can define spectrum as a beam of light that can be separated in the form of light waves or radio waves. The common example of visible spectrum is the rainbow in which we are able to see different colours separated by a thin boundary.
Now, let us find the line spectrum. We can define a line spectrum as a dark or bright line which results either from the emission or absorption of light in a narrow frequency range.
Now let us discuss how a line spectrum is produced. When there is an interaction between atoms, molecules and a photon of specific frequency, it creates a line spectrum in which the photon is absorbed by the electron and is detected by an observer as a separate line.
Line spectrum can be emitted from arc lamps or other systems in the form of electrical discharge in a sealed gas tube. Neon lamps, hollow cathode lamps, electrode less lamps are some of the examples through which line spectrum can be emitted.
As it can be seen in step two that the source of emission of line spectrum is a sealed gas tube, so from the given options, mercury lamp satisfies this criteria.
Hence, option (C), i.e. Mercury lamp is the correct choice for the given question.
Note: We should not get confused with the filament of the bulb, as it is just a part of the bulb. If in the option it would have been given only a bulb, in that case it can be considered as a source of line spectrum emission. The spectrum of different wavelengths are classified as different bands and known as electromagnetic spectrum range.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

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

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

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Physics Average Value and RMS Value JEE Main 2025

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

Displacement-Time Graph and Velocity-Time Graph for JEE
