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

There are two sources of light, each emitting with a power of $100 \,W.$ One emits $X-$ rays of wavelength $1\, nm$ and the other visible light at $500\, nm$. Find the ratio of number of photons of $X-$ rays to the photons of visible light of the given wavelength ? 

A $10\, kW$ transmitter emits radio waves of wavelength $500\, m$. The number of photons emitted per second by the transmitter is of the order of

The number of photons per second on an average emitted by the source of monochromatic light of wavelength $600\, \mathrm{~nm}$, when it delivers the power of $3.3 \times 10^{-3}$ $watt$ will be : $\left(\mathrm{h}=6.6 \times 10^{-34}\, \mathrm{Js}\right)$

Photo-electric effect can be explained by

In a photoemissive cell with executing wavelength $\lambda $, the fastest electron has speed $v.$ If the exciting wavelength is changed to $\frac{{3\lambda }}{4}$, the speed of the fastest emitted electron will be