Home
Class 12
PHYSICS
When a particle accelerated with 200 V p...

When a particle accelerated with 200 V potential its de Broglie wavelength is `10^(-10)`m. Find the wavelength (in Å) of the same particle if it is accelerated by 800 V.

Text Solution

Verified by Experts

The correct Answer is:
`00.50`
By using `lambda.sigma(1)/(sqrt(V))`
`(lambda_(1))/(lambda_(2))= sqrt((v_(2))/(V_(1)))`
`:. (10^(-10))/(lambda_(2))=sqrt((800)/(200))=2`
`:.lambda_(2)=0.5A`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • NTA TPC JEE MAIN TEST 55

    NTA MOCK TESTS|Exercise PHYSICS (SUBJECTIVE NUMERICAL)|10 Videos

  • NTA TPC JEE MAIN TEST 54

    NTA MOCK TESTS|Exercise PHYSICS|30 Videos

  • NTA TPC JEE MAIN TEST 56

    NTA MOCK TESTS|Exercise PHYSICS |30 Videos

Similar Questions

Explore conceptually related problems

The de-Broglie wavelength associated with a material particlewhen it is accelerated through a potential difference of 150volt is 1A. The de-Broglie wavelength associated with the same particle when it is accelerated through a potential difference of 1350 V will be:

An electron is accelerated through a potential of 120 V. Find its de Broglie wavelength

Knowledge Check

  • A proton is accelerated through 225 V . Its de Broglie wavelength is

    A
    0.1 nm
    B
    0.2 nm
    C
    0.3 nm
    D
    0.4nm

  • A proton is accelerated through 225 V . Its de Broglie wavelength is

    A
    0.001 nm
    B
    0.2 nm
    C
    0.3 nm
    D
    0.4 nm

  • For a proton accelerated through V volts , de Broglie wavelength is given as lambda =

    A
    `(12.27)/(sqrt(V)) Å`
    B
    `(0.101)/(sqrt(V)) Å `
    C
    ` (0.286 )/(sqrt(V)) Å `
    D
    `(12400)/(V) Å `

    • Similar Questions

    Explore conceptually related problems

    An alpha - particle is accelerated through a potential difference of 100 V. Its de-Broglie's wavelength is

    An electron of mass m when accelerated through a potential difference V, has de Broglie wavelength lambda . The de Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be :-

    An a particle and a proton are accelerated by same potential . Find ratio of their de Broglie wavelengths.

    A proton accelerated through a potential difference of 100V, has de-Broglie wavelength lamda_0 . The de-Broglie wavelength of an alpha -particle, accelerated through 800V is

    A proton accelerated through a potential V has de-Broglie wavelength lamda . Then, the de-Broglie wavelengthof an alpha-particle, when accelerated through the same potential V is

    Home
    Profile