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)

The stopping potential V₀ (in volt) as a function of frequency ($$\upsilon $$) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck’s constant (h) = 6.63 × 10^–34 Js, electron charges e = 1.6 × 10^–19 C)

A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck's constant h = 6.6 × 10^–34 Js, speed of light c = 3.0 × 10⁸ m/s]