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 $c = 3 \times 10^8 \ m/s$):

Find the number of photons emitted per second by a $10 \ kW$ radio transmitter operating at a wavelength of $500 \ m$.

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

The curve drawn between the velocity and frequency of a photon in a vacuum will be a

If the energy of a photon corresponding to a wavelength of $6000 \mathring{A}$ is $3.2 \times 10^{-19} \ J$,the photon energy for a wavelength of $4000 \mathring{A}$ will be . . . . . . .

An atom absorbs a photon of wavelength $500\,nm$ and emits another photon of wavelength $600\,nm$. The net energy absorbed by the atom in this process is $n \times 10^{-4}\,eV$. The value of $n$ is ............ [Assume the atom to be stationary during the absorption and emission process] (Take $h = 6.6 \times 10^{-34}\,Js$ and $c = 3 \times 10^8\,m/s$)