Dual Nature of Radiation and Matter
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4 months agoContributor-Level 10
11.5 The energy flux of sunlight, = 1.388 W/
Hence power of the sunlight per square meter, P = 1.388
Speed of light, c = 3 m/s
Planck's constant, h = 6.626 Js
Average wavelength of photon, = 550 nm = 550 m
If n is the number of photon per square meter, incident on earth per second, the equation of power can be written as
or =
We know =
Hence, = = = 3.84 photons /s
Therefore, every second 3.84 photons are incident per square meter on earth.
New answer posted
4 months agoContributor-Level 10
11.4 Wavelength of the monochromatic light, = 632.8 m
Power emitted by laser, P = 9.42 mW = 9.42 W
Planck's constant, h = 6.626 Js
Speed of light, c = 3 m/s
Mass of hydrogen atom, m = 1.66 kg
The energy of each photon is given as, = = J = 3.141 J
The momentum of each photon is given by = = = 1.047 kg-m/s
Number of photons arriving per second at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power c
New answer posted
4 months agoContributor-Level 10
11.3 Photoelectric cut-off voltage, = 1.5 V
Maximum kinetic energy of the photoelectrons emitted is given as
, where e = charge of an electron = 1.6 C
K = 1.6 = 2.4 J
Hence, maximum kinetic energy of the photoelectrons emitted is 2.4 .
New answer posted
4 months agoContributor-Level 10
11.2 Work function of cesium metal, = 2.14 eV
Frequency of light, = 6 Hz
The maximum kinetic energy is given by the photoelectric effect, = ,
Where = Planck's constant = 6.626 Js, 1 eV = 1.602 J
= = 0.345 eV
For stopping potential , we can write the equation for kinetic energy as:
= , where e = charge of an electron =
or = = 0.345 V
Hence, the stopping potential is 0.345 V
Maximum speed of the emitted photoelectrons = v
Kinetic energy K = m , where m = mass of the e
New answer posted
4 months agoContributor-Level 10
11.1 Potential of the electrons, V = 30 kV = 3 V
Energy of the electron, E = e where e = charge of an electron = 1.6
(a) Maximum frequency produced by the X-ray =
The energy of the electron is given by the relation, E = h ,
where h = Planck's constant = 6.626 Js
= = 7.244 Hz
7.244 Hz
(b) The minimum wavelength produced is given as
where c = Speed of light in air, c = 3 m/s
= 4.14 m = 0.0414 nm
Hence, the minimum wavelength produced is 0.414 nm.
New answer posted
5 months agoBeginner-Level 5
The Davisson-Germer experiment was conducted to test the De-broglie hypothesis and know wave nature of electrons. In this experiment, a beam of electrons was emitted from an electron gun with known acceleration to strike a nickel crystal placed inside a vacuum chamber.
After striking the surface, electrons are scattered from the crystal surface at different angles. The diffraction pattern obtained through this experiment was similar to that produced by X-rays, which is wave.
The Davisson-Germer experiment provided the first experimental proof of the wave nature of matter and de Broglie's hypothesis right.
New answer posted
5 months agoBeginner-Level 5
Significance of de Broglie's Hypothesis in Modern Physics (NCERT-based answer):
De Broglie's hypothesis is a revolutionary idea which changed the understanding of matter and wave as we know it. This hypothesis introduced the concept that particles of matter, like electrons, and Proton, can behave like wave and have wave-like properties, just as light exhibits both wave and particle nature.
According to de Broglie, any moving particle has a wavelength given by:
where is the wavelength, is Planck's constant, is the mass, and is the velocity of the particle.
New answer posted
5 months agoBeginner-Level 5
The photoelectric effect when light is projected in a metal surface, it's surface ejects electron (which are in outermost shell or free electron). This happens because the light provides additional energy to the electron to detach itself from metal surface. Well to do this light must have a certain amount of energy which in other words can be said the right frequency of light is required for photoelectric effect.
This frequency must be above a certain minimum value (called threshold frequency) fixed for every metal, regardless of its intensity. Photoelectric effect cannot be explained by the wave theory of light.
New answer posted
6 months agoContributor-Level 10
Students can download the NCERT Solutions PDF for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter from reliable educational platforms. However, candidates can also visit the homepage of Shiksha and click on the “NCERT Chemistry Class 12th Solution PDF - Dual Nature of Radiation and Matter Chapter Download” to donwload the PDF. These NCERT PDF Solutions are helpful to prepare without an internet connection. The Class 12 Physics NCERT Solutions consists of detailed explanations, formulas, and solved examples to help students prepare effectively.
New answer posted
6 months agoContributor-Level 10
NCERT Solutions for Class 12 Ch 11 Dual Nature of Radiation and Matter includes step-by-step explanations of numerical problems, theoretical concepts and derivations.
These Class 12 Ch 11 Dual Nature of Radiation and Matter NCERT Solutions help students not only to clear thier doubts but also useful to practice problem-solving, and to understand important topic and concepts that are frequently asked in CBSE board exams and competitive exams. Practicing with Class 12 Physics NCERT solutions PDFs improve conceptual clarity and boosts student's confidence for exams.
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