If the de Broglie wavelength of an electron is equal to the 10^–3 times the wavelength of a photon of frequency 6 $$ \times $$ 10¹⁴ Hz, then the speed of electron is equal to : (Speed of light = 3 $$ \times $$ 10⁸ m/s, Planck's constant = 6.63 $$ \times $$ 10^–34 J.s, Mass of electron = 9.1 $$ \times $$ 10^–31 kg)

A metal plate of area 1 $$ \times $$ 10^–4 m² is illuminated by a radiation of intensity 16 mW/m². The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photoelectrons per second and their maximum energy, respectively, will be

In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10^–12 m, the minimum electron energy required is close to -

The magnetic field associated with a light wave is given, at the origin, by B = B₀ [sin(3.14 $$ \times $$ 10⁷)ct + sin(6.28 $$ \times $$ 10⁷)ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(Take c = 3 $$ \times $$ 10⁸ ms^$$-$$1, h = 6.6 $$ \times $$ 10^$$-$$34J-s)