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Calculate the (a) momentum and (b) de-Br...

Calculate the (a) momentum and (b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V. Given, `h=6.63xx10^(-34)Js, m_(e)=9.1xx10^(-31)kg, e=1.6xx10^(-19)C`.

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To solve the question, we need to calculate two things: (a) the momentum of the electron and (b) the de-Broglie wavelength of the electron accelerated through a potential difference of 56V. ### Step-by-Step Solution: **(a) Calculate the momentum of the electron:** 1. **Understand the relationship between kinetic energy and potential energy:** When an electron is accelerated through a potential difference \( V \), the kinetic energy gained by the electron is equal to the work done on it by the electric field. This can be expressed as: ...
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What is the (a) momentum (b) speed and (c) de-Broglie wavelength of an electron with kinetic energy of 120 eV. Given h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J .

An electron is accelerated through a potential difference of 10,000V . Its de Broglie wavelength is, (nearly): (me=9xx10^(-31)kg)

Knowledge Check

  • The de Broglie wavelength of an electron with kinetic energy 120 e V is ("Given h"=6.63xx10^(-34)Js,m_(e)=9xx10^(-31)kg,1e V=1.6xx10^(-19)J)

    A
    `2.13Ã…`
    B
    `1.13Ã…`
    C
    `4.15Ã…`
    D
    `3.14Ã…`

  • When light of wavelength 400nm is incident on the cathode of photocell, the stopping potential recorded is 6V. If the wavelength of the incident light is to 600nm, calculate the new stopping potential. [Given h=6.6xx10^(-34) Js, c=3xx10^(8)m//s , e=1.6xx10^(-19)C ]

    A
    `1.03 V`
    B
    `2.42 V`
    C
    `4.97 V`
    D
    `3.58 V`

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    Explore conceptually related problems

    If de Broglie wavelength of an electron is 0.5467 Å, find the kinetic energy of electron in eV. Given h=6.6xx10^(-34) Js , e= 1.6xx10^(-19) C, m_e=9.11xx10^(-31) kg.

    De - Broglie wavelength of an electron accelerated by a voltage of 50 V is close to (|e|=1.6xx10^(-19)C,m_(e)=9.1xx10^(-31))

    Find de-Broglie wavelength of electron with KE =9.6xx10^(-19)J .

    If an electron has an energy such that its de Broglie wavelength is 5500 Å , then the energy value of that electron is (h= 6.6 xx 10^(-34)) Js, m_( c) = 9.1 xx 10^(-31) kg

    With what velocity must an electron travel so that its momentum is equal to that of a photon with a wavelength of 5000Å(h=6.6 xx 10^(-34)Js,m_(e)=9.1 xx 10^(-31) Kg)

    Wave property of electron implies that they will show diffraction effected . Davisson and Germer demonstrated this by diffracting electron from crystals . The law governing the diffraction from a crystals is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructiely Electron accelerated by potential V are diffracted from a crystal if d = 1 Å and i = 30^(@), V should be about (h = 6.6 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg , e = 1.6 xx 10^(-19)C)

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