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)

Surface of certain metal is first illuminated with light of wavelength $$\lambda $$₁ = 350 nm and then, by light of wavelength $$\lambda $$₂ = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :

(Energy of photon n = $${{1240} \over {\lambda (in\,mm)}}$$eV)