The momentum of a photon with energy $20 \, eV$ is

The principle of $LASER$ action involves:

$A$ source $S_1$ is producing $10^{15}$ photons per second of wavelength $5000 \; \mathring{A}$. Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100 \; \mathring{A}$. Then,$(\text{power of } S_2) / (\text{power of } S_1)$ is equal to

The ratio of energies of photons with wavelengths $\lambda_1 = 150 \ nm$ and $\lambda_2 = 300 \ nm$ is:

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

An electric bulb of $100 \ W$ power emits photons of wavelength $410 \ nm$ per second. The number of photons emitted per second is: $(h = 6.63 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s)$