Home
Class 12
PHYSICS
When photons of energy 4.25 eV strike th...

When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, then

A

The work function of A is 2.25 eV

B

The work function of B is `4.20 eV`

C

`T_(A) = 2.00 eV`

D

`T_(B) = 2.75 eV`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
(1) `lambda_(A) = (h)/(sqrt(2mT_(A)))` , (2) `lambda_(B) = (h)/(sqrt(2mT_(B)))`
(3) `T_(B) = T_(A) - 1.5 eV` , (4) `lambda_(B) = 2lambda_(A)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Exercise - 10|2 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Exercise - 11|2 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN |Exercise Exercise - 08|2 Videos

  • RACE

    ALLEN |Exercise Basic Maths (Wave Motion & Dopplers Effect) (Stationary waves & doppler effect, beats)|25 Videos

  • TEST PAPER

    ALLEN |Exercise PHYSICS|4 Videos

Similar Questions

Explore conceptually related problems

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV . The stopping potential , in volt is

An electron, in a hydrogen like atom , is in excited state. It has a total energy of -3.4 eV , find the de-Broglie wavelength of the electron.

Knowledge Check

  • What is the de-Broglie wavelength of a electron with kinetic energy of 120 eV?

    A
    `1121xx10^(-13)m`
    B
    `1121xx10^(-12)m`
    C
    `0.1121xx10^(-11)m`
    D
    `0.1121xx10^(-10)m`

  • Two photons of energy 4 eV and 4.5 eV are incident on two metals A and B respectively.The maximum kinetic energy for an ejected electron is T_(A) for A and T_(B)=T_(A)-1.5 eV for the metal B.The relation between the de-Broglie wavelengths of the ejected electron of A and B are lambda_(B)=2lambda_(B) .The work function of the metal B is

    A
    1.5 eV
    B
    3 eV
    C
    4eV
    D
    4.5 eV

  • When radiation of 400 nm wavelength Is incident on photocell ,emitter produces photoelectron with maximum kinetic energy of 1.68 eV work function of emitter will be………. [hc=1240 eVnm]

    A
    3.09 eV
    B
    1.42 eV
    C
    1.51 eV
    D
    1.68 V

    • Similar Questions

    Explore conceptually related problems

    Photons with energy 5 eV are incident on a cathode C in a photoelectric cell . The maximum energy of emitted photoelectrons is 2 eV . When photons of energy 6 eV are incident on C , no photoelectrons will reach the anode A , if the stopping potential of A relative to C is

    In an excited state of hydrogen like atom an electron has total energy of -3.4 eV . If the kinetic energy of the electron is E and its de-Broglie wavelength is lambda , then

    An electron with initial kinetic energy of 100 eV is acclerated through a potential difference of 50 V . Now the de-Brogile wavelength of electron becomes

    What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.

    What is the (a)Momentum (b)speed ,and (c )de-Broglie wavelength of an electron with kinetic energy of 120 eV.

    Home
    Profile