JEE Exam > JEE Questions > The atomic masses of He and Ne are 4 and 20 a...
Start Learning for Free
The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73°C is "M" times that of the de-Broglie wavelength of Ne at 727°C. M is
(2013 Adv., Integer Type)
Correct answer is '5'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
The atomic masses of He and Ne are 4 and 20 amu, respectively. The val...
Most Upvoted Answer
The atomic masses of He and Ne are 4 and 20 amu, respectively. The val...
De Broglie Wavelength
The de Broglie wavelength is a concept in quantum mechanics that describes the wave-like nature of particles. It is given by the equation:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the particle.
De Broglie Wavelength and Kinetic Energy
The momentum of a particle can be related to its kinetic energy using the equation:
p = √(2mK)
where p is the momentum, m is the mass of the particle, and K is the kinetic energy.
Relation between De Broglie Wavelength and Temperature
The kinetic energy of a gas particle is directly proportional to its temperature. Therefore, the de Broglie wavelength of a gas particle can be related to its temperature.
Relation between De Broglie Wavelength and Atomic Mass
The de Broglie wavelength is inversely proportional to the square root of the mass of a particle. Therefore, the de Broglie wavelength of a heavier particle will be smaller than that of a lighter particle.
Calculating the Value of M
Given:
Mass of He (m1) = 4 amu
Mass of Ne (m2) = 20 amu
Temperature of He (T1) = -73°C = -73 + 273 = 200 K
Temperature of Ne (T2) = 727°C = 727 + 273 = 1000 K
Using the relation between de Broglie wavelength and temperature, we can write:
λ1 / λ2 = √(T2 / T1)
Substituting the given values:
λ1 / λ2 = √(1000 / 200) = √5
Using the relation between de Broglie wavelength and atomic mass, we can write:
λ1 / λ2 = √(m2 / m1)
Substituting the given values:
√5 = √(20 / 4) = √5
Therefore, M = 5.
The de Broglie wavelength is a concept in quantum mechanics that describes the wave-like nature of particles. It is given by the equation:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 J·s), and p is the momentum of the particle.
De Broglie Wavelength and Kinetic Energy
The momentum of a particle can be related to its kinetic energy using the equation:
p = √(2mK)
where p is the momentum, m is the mass of the particle, and K is the kinetic energy.
Relation between De Broglie Wavelength and Temperature
The kinetic energy of a gas particle is directly proportional to its temperature. Therefore, the de Broglie wavelength of a gas particle can be related to its temperature.
Relation between De Broglie Wavelength and Atomic Mass
The de Broglie wavelength is inversely proportional to the square root of the mass of a particle. Therefore, the de Broglie wavelength of a heavier particle will be smaller than that of a lighter particle.
Calculating the Value of M
Given:
Mass of He (m1) = 4 amu
Mass of Ne (m2) = 20 amu
Temperature of He (T1) = -73°C = -73 + 273 = 200 K
Temperature of Ne (T2) = 727°C = 727 + 273 = 1000 K
Using the relation between de Broglie wavelength and temperature, we can write:
λ1 / λ2 = √(T2 / T1)
Substituting the given values:
λ1 / λ2 = √(1000 / 200) = √5
Using the relation between de Broglie wavelength and atomic mass, we can write:
λ1 / λ2 = √(m2 / m1)
Substituting the given values:
√5 = √(20 / 4) = √5
Therefore, M = 5.
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer?
Question Description
The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer?.
The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer?.
Solutions for The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer?, a detailed solution for The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? has been provided alongside types of The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The atomic masses of He and Ne are 4 and 20 amu, respectively. The value of the de-Broglie wavelength of He gas at -73C is M times that of the de-Broglie wavelength of Ne at 727C. M is(2013 Adv., Integer Type)Correct answer is '5'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup