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If the de Broglie wavelength of an electron is equal to  10-3 times the wavelength of a photon of frequency  6×1014 then the speed (in m/s) of electron is equal to: (Speed of light =3×108 m/s) Planck's constant =
6.63×10-34 J.s Mass of electron =9.1×10-31 kg) is x × 104.what is x
    Correct answer is '145'. Can you explain this answer?
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    To solve this problem, we can use the de Broglie wavelength formula:

    λ = h / p

    where λ is the de Broglie wavelength, h is the Planck's constant (6.626 x 10^-34 J*s), and p is the momentum.

    First, let's find the momentum of the photon using the formula:

    E = hf

    where E is the energy, h is the Planck's constant, and f is the frequency.

    Since the energy of a photon is given by E = pc, where c is the speed of light (3 x 10^8 m/s), we can equate the two formulas:

    hf = pc

    Simplifying, we get:

    p = h / λ

    Now, we can substitute the given values into the formulas:

    λ_photon = c / f_photon

    Given: f_photon = 6 Hz

    λ_photon = 3 x 10^8 m/s / 6 Hz = 5 x 10^7 m

    λ_electron = 10^-3 * λ_photon = 10^-3 * 5 x 10^7 m = 5 x 10^4 m

    Now, we can calculate the momentum of the electron:

    p_electron = h / λ_electron

    p_electron = 6.626 x 10^-34 J*s / 5 x 10^4 m = 1.3252 x 10^-38 J*s/m

    Therefore, the momentum of the electron is approximately 1.3252 x 10^-38 J*s/m.
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    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

    Can you explain the answer of this question below: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 correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving with same kinetic energy is :A:H D TB:H = D = TC:H D TD:H D TThe answer is a.

    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. If the uncertainty in velocity position is same, then the uncertainty in momentum will be

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    If the de Broglie wavelength of an electron is equal to10-3times the wavelength of a photon of frequency 6×1014then the speed (in m/s) of electron is equal to: (Speed of light=3×108m/s) Plancks constant=6.63×10-34J.sMass of electron=9.1×10-31kg) is x× 104.what is xCorrect answer is '145'. Can you explain this answer?
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    If the de Broglie wavelength of an electron is equal to10-3times the wavelength of a photon of frequency 6×1014then the speed (in m/s) of electron is equal to: (Speed of light=3×108m/s) Plancks constant=6.63×10-34J.sMass of electron=9.1×10-31kg) is x× 104.what is xCorrect answer is '145'. 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 If the de Broglie wavelength of an electron is equal to10-3times the wavelength of a photon of frequency 6×1014then the speed (in m/s) of electron is equal to: (Speed of light=3×108m/s) Plancks constant=6.63×10-34J.sMass of electron=9.1×10-31kg) is x× 104.what is xCorrect answer is '145'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the de Broglie wavelength of an electron is equal to10-3times the wavelength of a photon of frequency 6×1014then the speed (in m/s) of electron is equal to: (Speed of light=3×108m/s) Plancks constant=6.63×10-34J.sMass of electron=9.1×10-31kg) is x× 104.what is xCorrect answer is '145'. Can you explain this answer?.
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