Show that for an electron laccelerated by a potential difference of V- volts, the de - Broglie wavelength will be `12.27/sqrtv overset@A`.

Statement I: If a stationary electron is accelerated with a potential difference of 1V, its de Broglie wavelength becomes 12.27 Å approximately. Statement II: The relation between the de Broglie wavelength lamda and the accelerating potential V of an electron is given by lamda = (12.27)/(V) Å

Determine de Broglie wavelength of an electron accelerated through a potential difference of V volt.

An electron of mass 'm' is accelerated by a potential difference V and the corresponding de-Broglie wavelength is lamda The de Broglie wavelength of a proton of mass M if it is accelerated by the same potential difference is.

An electron is accelerated by a photential difference V and as a result, its de Broglie wavelength beocmes lamda . If the applied potential difference was 2V, the de Broglie wavelength would have been

An electron (charge = e, mass = m) is accelerated from rest through a potential difference of V volt. What will be the de Broglie wavelength of the electron?

The de Broglie wavelength associated with an electron accelerated through a potential difference V is lamda . What will be its wavelength when the accelerating potential is increased to 4 V?

An alpha -particle and a proton are accelerated through the same potential difference. Find the ratio of de Broglie wavelengths.

A proton and an electron have the same kinetic energy. Which one has large de Broglie wavelength?

Calculate de Broglie wavelength of an electron beam accelerated through a potential difference of 60V.

An electron accelerated through a potential of 10000 V from rest has a de Broglie wavelength lamda . What should be the accelerating potential so that the wavelength is doubled?