Home
Class 12
PHYSICS
Electrons are accelerated through a p.d....

Electrons are accelerated through a `p.d`. Of `150V`. Given `m=9.1xx10^(-31)kg,e=1.6xx10^(-19)c,h=6.62xx10^(-34)Js`, the de Broglie wavelength associated with it is

A

`1.5 A^(@)`

B

`1.0 A^(@)`

C

`3.0 A^(@)`

D

`0.5 A^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B

`lambda=sqrt(150/V)`
Promotional Banner

Topper's Solved these Questions

  • DUAL NATURE

    NARAYNA|Exercise LEVEL-II (H.W)|17 Videos
  • DUAL NATURE

    NARAYNA|Exercise LEVEL-I(H.W)|7 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    NARAYNA|Exercise ASSERTION AND REASON|15 Videos
  • ELECTRIC CHARGES AND FIELDS

    NARAYNA|Exercise EXERCISE -4|43 Videos

Similar Questions

Explore conceptually related problems

Photon having energy equivalent to the binding energy of 4th state of He^(+) ion is used to eject an electron from the metal with KE/2eV . If electron is further accelerated through a potential difference of 4V then the minimum value of de Broglie wavelength associated with the electron is : (h = 6.6 xx 10^(-34) J-s , m_(e) = 9.1 xx 10^(-31) kg . 1 eV = 1.6 xx 10^(-19) J )

Calculate the de-Broglie wavelength of an electron of kinetic energy 100 eV. Given m_(e)=9.1xx10^(-31)kg, h=6.62xx10^(-34)Js .

Given m_(e)=9.11 xx 10^(-31)kg and h = 6.626 xx 10^(-34)Js , the uncertainty involved in the measuremenetof velocity within a distance of 0.1Å is

If the velocity of the electron in Bohr's first orbit is 2.19xx10^(6) m s^(-1) , calculate the de Broglie wavelength associated with it.

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 .

What is the wavelength of light . Given energy =3.03xx10^(-19)J, h=6.6xx10^(-34)JS, c=3.0xx10^(8)m//s ?

Calculate the wavelength of an electron that has been accelerated in a particle accelerator through a potentiation difference of 100 million volts [ 1eV = 1.6 xx 10^(-19)J, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) J s, c = 3.0 xx 10^(8) m s^(-1)]

An electron of mass 9.1xx10^(-31) kg and charge 1.6xx10^(-19) C is accelarted through a potential difference of V volt. The de Broglie wavelength (lambda) associated with the electron is

Calculate the wavelength of an electron that has been accelerated in a particle acceleratior through a potential difference of 100 million volts. (1 eV = 1.6 xx 10^(-19)C, m_e = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) Js) .