The number of photons of wavelength $540 \ nm$ emitted per second by an electric bulb of power $100 \ W$ is (taking $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$):

Calculate the energy of a photon having wavelength $\lambda = 1000 \, \mathring{A}$ in Joules.

An electron and a proton are detected in a cosmic ray experiment,the first with kinetic energy $10 \; keV$,and the second with $100 \; keV$. Which is faster,the electron or the proton? Obtain the ratio of their speeds. (Electron mass $= 9.11 \times 10^{-31} \; kg$,proton mass $= 1.67 \times 10^{-27} \; kg$,$1 \; eV = 1.60 \times 10^{-19} \; J$)

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

When light of wavelength $\lambda$ is incident on a photosensitive surface,photons of power $P$ are emitted. The number of photons $n$ emitted in time $t$ is: [$h$ = Planck's constant,$c$ = velocity of light in vacuum]

If the momentum of a photon is $p$,then its frequency is ........