All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
JEE Exam  >  JEE Questions  >  The value of the de-Broglie wavelength of He ... Start Learning for Free
SaveJoin the Discussion on the App
The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?
    Correct answer is '8'. Can you explain this answer?
    FREE This question is part of
    Download PDF Attempt Test
    		Download PDF Attempt this Test
    Verified Answer
    The value of the de-Broglie wavelength of He gas at 100 K is 5 times t...


    View all questions of this test
    Most Upvoted Answer
    The value of the de-Broglie wavelength of He gas at 100 K is 5 times t...
    Given Data:
    - Temperature of He gas (T1) = 100 K
    - Temperature of gas X (T2) = 1250 K
    - Ratio of de Broglie wavelength of He gas to that of gas X = 5

    Formula:
    The de Broglie wavelength (λ) is given by the equation:
    λ = h / (mv)

    Where:
    λ = de Broglie wavelength
    h = Planck's constant (6.626 x 10^-34 Js)
    m = mass of the particle
    v = velocity of the particle

    Step 1: Calculate the de Broglie wavelength of He gas
    Let the atomic mass of He gas be M1.

    Using the given formula, we have:
    λ1 = h / (m1v1)

    Step 2: Calculate the de Broglie wavelength of gas X
    Let the atomic mass of gas X be M2.

    Using the given formula, we have:
    λ2 = h / (m2v2)

    Step 3: Calculate the ratio of de Broglie wavelengths
    Given: λ1 / λ2 = 5

    Substituting the values from Step 1 and Step 2, we get:
    (h / (m1v1)) / (h / (m2v2)) = 5

    Simplifying the expression, we get:
    (m2v2) / (m1v1) = 5

    Step 4: Calculate the ratio of velocities
    Since we know that the de Broglie wavelength is inversely proportional to velocity, we can write:
    v2 / v1 = λ1 / λ2 = 1/5

    Step 5: Calculate the ratio of masses
    Using the kinetic energy equation:
    1/2 m1 v1^2 = 1/2 m2 v2^2

    Substituting the values from Step 4, we get:
    1/2 m1 v1^2 = 1/2 m2 (5v1)^2
    m1 v1^2 = 25 m2 v1^2
    m2 / m1 = 1/25

    Step 6: Calculate the atomic mass of X
    Since the ratio of masses is given as 1/25, and we know that the atomic mass of He is 4, we can write:
    4 / m1 = 1/25
    m1 = 4*25 = 100
    The atomic mass of X (m2) is therefore 100 - 4 = 96

    Final Answer:
    The atomic mass of X is 96.
    < Previous Question Next Question >
    Explore Courses for JEE exam

    Similar JEE Doubts

    The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. The transition, so that the de - Broglie wavelength of electron becomes 3 times of its initial value in He+ ion will be

    Top Courses for JEEView all

    Top Courses for JEE

    The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer?
    Question Description
    The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. 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 value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. 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 value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer?.
    Solutions for The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. 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 value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer?, a detailed solution for The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer? has been provided alongside types of The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The value of the de-Broglie wavelength of He gas at 100 K is 5 times that of the de-Broglie wavelength gas X, at 1250 K. What is atomic mass of X?Correct answer is '8'. Can you explain this answer? tests, examples and also practice JEE tests.
    Explore Courses for JEE exam

    Top Courses for JEE

    Explore Courses

    Suggested Free Tests

    Signup for Free!
    Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
    10M+ students study on EduRev
    Get App
    © 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