In the Compton scattering process,the incident $X$-radiation is scattered at an angle of $60^{\circ}$. The wavelength of the scattered radiation is $0.22 \ \text{Å}$. The wavelength of the incident $X$-radiation in $\text{Å}$ units is:

The energy of a photon whose frequency is $10^{12} \text{ MHz}$ is:

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

$A$ $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu m$. Assuming it to be $25\%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is:

The ratio of the power of a light source $S_1$ to that of the light source $S_2$ is $2$. $S_1$ is emitting $2 \times 10^{15}$ photons per second at $600 \ nm$. If the wavelength of the source $S_2$ is $300 \ nm$,then the number of photons per second emitted by $S_2$ is . . . . . . $\times 10^{14}$.

Number of photons of equal energy emitted per second by a $6 \ mW$ laser source operating at $663 \ nm$ is . . . . . . . (Given: $h=6.63\times10^{-34} \ J.s$ and $c=3\times10^{8} \ m/s$)