Home
Class 12
PHYSICS
The wavelength of photon of lambda is eq...

The wavelength of photon of `lambda` is equal to the de-Broglie wave length of electron. Show that energy of photon `=2lambdamc/hxx` kinetic energy of electron .

Answer

Step by step text solution for The wavelength of photon of lambda is equal to the de-Broglie wave length of electron. Show that energy of photon =2lambdamc/hxx kinetic energy of electron . by PHYSICS experts to help you in doubts & scoring excellent marks in Class 12 exams.

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MODEL PAPER SET-01

    UNITED BOOK HOUSE|Exercise EXERCISE|40 Videos

  • MODEL PAPER SET-03

    UNITED BOOK HOUSE|Exercise EXERCISE|43 Videos

Similar Questions

Explore conceptually related problems

Calculate the de Broglie wavelength of an electron of kinetic energy 500 eV.

The wavelength lamda of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of the photon is (2 lamda mc)/(h) times of the kinetic energy of the electron. Here m, c and h have their usual meaning.

Knowledge Check

  • The de Broglie wavelength of an electron is the same as that 50 keV X-ray photon. The ratio of the energy of the photon to the kinetic energy of the electron is (the energy equivalent of electron mass is 0.5 MeV)

    A
    `1 : 50`
    B
    `1 : 20`
    C
    `20 : 1`
    D
    `50 :1`

    • Similar Questions

    Explore conceptually related problems

    If wavelength (lamda) of a photon and de Broglie wavelength of an electron are same, then show that the energy of the photon is (2 lamda mc)/(h) times the kinetic energy of the electron. Here, m, c and h have their usual meanings

    Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

    The energy of a photon of wavelength lambda is—

    Calculate the de Broglie Wavelength of an electron having kinetic energy 3.6 MeV.

    A photon and a proton have the same de – Broglie wavelength λ.Show that, the photon has energy 2mc lambda/h times the energy of proton.

    Show that the de Broglie wavelength of electron of energy K is given by lamda = (h)/(sqrt(2mK)) .

    A photon and an electron have same de Broglie wavelength. Which one has greater total energy?

    Home
    Profile