Calculate `lamda_(dB)` of electron have K.E. 3eV

The total energy of an electron in second excited state of hydrogen atom is -1.51eV. Calculate (i) KE of electron (ii) PE of electron in this state.

Calculate de Broglie wavelength of an electron having kinetic energy 120 eV.

The wavelength lamda_(e) of ann electron and lamda_(p) of a photon of same energy E are related by

What will be the de - Broglie wavelength of an electron having K.E of 500 eV?

Work function of a cathode is 4.2 eV. Energy of incident photons is 6.2 eV . Anode is kept at + 2 volt w.r.t cathode. Calculate minimum and maximum K.E. of electrons reaching at the anode.

Electron are accelerated through a potential difference V and protons are accelerated through a potential difference of 4 V. The de-Broglie wavelength are lamda_e and lamda_p for electrons and protons, respectively The ratio of lamda_e/lamda_p is given by ( given , m_e is mass of electron and m_p is mass of proton )

Electron are accelerated through a potential difference V and protons are accelerated through a potential difference of 4 V. The de-Broglie wavelength are lamda_e and lamda_p for electrons and protons, respectively The ratio of lamda_e/lamda_p is given by ( given , m_e is mass of electron and m_p is mass of proton )