$A$ photon,an electron,and a uranium nucleus all have the same wavelength. The one with the most energy is:

$A$ light bulb emits $5\%$ of its energy as visible light. How many quanta are emitted per second by a $100 \ W$ lamp? (Assume the wavelength of visible light is $5.6 \times 10^{-5} \ cm$)

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never 'count photons',even in barely detectable light.
$(a)$ The number of photons emitted per second by a Medium wave transmitter of $10\; kW$ power,emitting radiowaves of wavelength $500\; m$.
$(b)$ The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive $(10^{-10}\; W m^{-2})$. Take the area of the pupil to be about $0.4\; cm^2$,and the average frequency of white light to be about $6 \times 10^{14}\; Hz$.

$A$ radio transmitter operates at a frequency of $880 \ kHz$ and a power of $10 \ kW$. The number of photons emitted per second is .......

Two sources of light emit $X$-rays of wavelength $1 \ nm$ and visible light of wavelength $500 \ nm$,respectively. Both sources emit light of the same power $200 \ W$. The ratio of the number of photons emitted per second by the $X$-ray source to the number of photons emitted per second by the visible light source is:

The principle of $LASER$ action involves: