If electrons are accelerated through a potential of `V` volt, the minimum wavelength of the continuous `X-ray` radiation is :

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 xx 10^(-2) nm, the potential difference is:

Wavelength of an electron accelerated through a potential difference of 1 volt is

Photon having energy equivalent to the binding energy of 4th state of He^(+) atom is used to eject an electron from the metal surface of work function 1.4 eV . If electrons are further accelerated through the potential difference of 4V then the minimum value of De-brogile wavelength associated with the electron is:

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

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

An electron is accelerated through a potential difference of 100 volts. What is the de-Broglie wavelength associated with it ? To which part of the electromagnetic does this value of wavelength correspond ?

The minimum wavelength lambda_min in the continuous spectrum of X-rays is

An electron is accelerated through a potential difference of V volt. It has a wavelength lamda associated must be accelerated so that its de Broglie wavelength is the same as the of a proton? Take mass of proton to be 1837 times larger than the mass of electron.

An electron is accelerated through a potential difference of 64volts. What is the de-broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond?

A proton when accelerated through a potential difference of V volt has a wavelength lamda associated with it. An alpha-particle in order to have the same lamda must be accelerated through a potential difference of