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)

Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths ^{$$\lambda $$}N, ^{$$\lambda $$}A respectively. The ratio $${{{}^\lambda N} \over {{}^\lambda A}}$$is closest to :

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 :