Find the number of electrons emitted per second by a $24 \,W$ source of monochromatic light of wavelength $6600 \mathring A$, assuming $3 \%$ efficiency for photoelectric effect (take $h=6.6 \times 10^{-34} \,Js$ ) $S$

The eye can detect  $5 ×10^4$ photons per square metre per sec of green light ($\lambda$ $= 5000\ \mathop A\limits^o $) while the ear can detect ${10^{ – 13}}\,(W/{m^2})$. The factor by which the eye is more sensitive as a power detector than the ear is close to

A  radiation of energy $'E'$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $( C =$ Velocity of light $)$

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths $\lambda _N,\,\lambda _A$  respectively. The ratio $\frac{{{\lambda _N}}}{{{\lambda _A}}}$  is closest to

According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is