A particle moving with kinetic energy E has de Broglie wavelength $$\lambda $$. If energy $$\Delta $$E is added to its energy, the wavelength become $$\lambda $$/2. Value of $$\Delta $$E, is :

An electron (mass m) with initial velocity $$\overrightarrow v = {v_0}\widehat i + {v_0}\widehat j$$ is in an electric field $$\overrightarrow E = - {E_0}\widehat k$$. If $$\lambda _0$$ is initial de-Broglie wavelength of electron, its de-Broglie wave length at time t is given by :

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_A eV end de-Broglie wavelength $$\lambda _A$$. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is T_B = (T_A – 1.5) eV. If the de-Broglie wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$, then the work function of metal B is :

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)