Show that, the relation between kinetic energy and the de-broglie wavelength associated with a moving electron is X = `h/(sqrt2mE`

Derive a relation between kinetic energy and de Broglie wavelength associated with a moving electron.

A particle is moving with kinetic energy E. It's de Broglie wavelength is—

The de Broglie wavelength of a moving electron is 1 Å . What is its velocity?

What happens to the de Broglie wavelength associated with a moving particle if its velocity is doubled?

Calculate the kinetic energy of a moving electron associated with a wavelength of 4.8 pm.

State the Heisenberg uncertainty principle. Calculate the de Broglie wavelength associated with an electron moving with velocity of I.0xxl0^7ms . (Mass of an electron : 9.1 xx 10^-31 kg ).

An electron and a proton are moving in the same direction and posses same kinetic energies. Find the ratio of de Broglie wavelength associated with these particles.

(a) Write two differences between electromagnetic wave and matter wave. (b) show that the circumference of the 'n' th electronic orbit is n times the de Broglie wavelength of the wave associated with the motion of electron i.e., 2pir_(n)=nlamda .

Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

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