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)

The de-Broglie wavelength ($$\lambda $$_B) associated with the electron orbiting in the second excited state of hydrogen atom is related to that in the ground state ($$\lambda $$_G) by :