Home
Class 12
PHYSICS
An electron with initial kinetic energy ...

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

A

`1 Å`

B

`sqrt(1.5Å)`

C

`sqrt(3)Å`

D

`12.27Å`

Text Solution

Verified by Experts

The correct Answer is:
A
For electron `lambda_(db)=sqrt(150/(100+50))=1 Å`
Promotional Banner

Topper's Solved these Questions

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise - 04[A]|25 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise-04 [B]|105 Videos

  • SIMPLE HARMONIC MOTION

    ALLEN|Exercise Exercise - 33|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

Electrons are accelerated through a potential difference of 150V . Calculate the de broglie wavelength.

Kinetic energy of an electron accelerated in a potential difference of 100 V is

The kinetic energy of an electron accelerated from rest through a potential difference of 5V will be

An electron is accelerated through a potential difference of V volit .Find th e de Broglie wavelength associated with electron.

An electron is accelerated by a potential difference of 50 V. Find the de Broglie wavelength associated with it.

An alpha - particle has initial kinetic energy of 25 eV and it is accelerated through a potential difference of 150 volt . If a photon has initial kinetic energy of 25 eV and it is accelerated through a potential difference of 25 volt , then find the approximate ratio of the final wavelengths associated with the proton and the alpha -particle .

An electron is accelerated through a potential difference of 100 V , then de-Broglie wavelength associated with it is approximately_____. A^(@)