An electron is accelerated under a potential difference of 64 V, the de-Brogile wavelength associated with electron is `[e = -1.6 xx 10^(-19) C, m_(e) = 9.1 xx 10^(-31)kg, h = 6.623 xx 10^(-34) Js]`

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 )

The kinetic energy of an electron is 5 eV . Calculate the de - Broglie wavelength associated with it ( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)

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) .

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)]

de-Brogile wavelength of an electron accelerated by a voltage of 50 V is close to ( |e| = 1.6 xx 10^(-19)C, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) Js) :-

de-Brogile wavelength of an electron accelerated by a voltage of 50 V is close to ( |e| = 1.6 xx 10^(-19)C, m_(e) = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) Js) :-

An electron is accelerated through a potential difference of 10,000 V.Its de-Broglie wavelength is ,(nearly) (m_(e)=9xx10^(-31)kg)

An electron is accelerated under a potential difference of 182 V. The maximum velocity of electron will be ........... (Charge of electron is 1.6 xx 10^(-19) C and its mass is 9.1 xx 10^(-31) kg )