Show.that the de-Broglie wavelength `lambda` of electrons of energy E is given by the relation, `lambda = h/sqrt(2mE)`.

Show that the de Broglie wavelength of electron of energy K is given by lamda = (h)/(sqrt(2mK)) .

Calculate the de Broglie wavelength of an electron of kinetic energy 500 eV.

What is the de Broglie wavelength related to an electron of energy 100 eV ? (Given, mass of electron, m = 9.1 xx 10^(-31) kg, " " e = 1.6 xx 10^(-19) C and h = 6.63 xx 10^(-34) J.s )

Proof that, the de Broglie, wavelength for an electron moving through the potential difference of V is given by lambda = h/sqrt(2meV) . From this equation show that for electron its value is. lambda= 12.27/sqrtV overset@A

Calculate the de Broglie Wavelength of an electron having kinetic energy 3.6 MeV.

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

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

Show that the de Broglie wavelength of charged particles accelerated through a potential of V volt is given by lamda = h//sqrt(2mqV) . Hence show that the de-Broglie wavelength of electron is (12.27)/(sqrtV)Å .

Calculate the de Broglie wavelength of an electron moving with a speed that is 1% of the speed of light.

The wavelength of photon of lambda is equal to the de-Broglie wave length of electron. Show that energy of photon =2lambdamc/hxx kinetic energy of electron .