The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is $$x \times 10^{13}$$. The value of x is _________. (Nearest integer)
(h = $$6.63 \times 10^{-34}$$ Js, c = $$3.00 \times 10^8$$ ms$$^{-1}$$)

We have been given a monochromatic infrared source whose power is $$P = 1\ \text{mW}$$. First, we convert this power into SI units. Since $$1\ \text{mW} = 10^{-3}\ \text{W}$$, we can write $$P = 1 \times 10^{-3}\ \text{W}$$.

The device is operated for a time interval of $$t = 0.1\ \text{s}$$. In physics, the total energy $$E_{\text{total}}$$ radiated in a given time is obtained from the very definition of power, which states

$$P = \frac{\text{Energy}}{\text{Time}} \; \Longrightarrow \; \text{Energy} = P \times t.$$

Substituting the numerical values, we get

$$E_{\text{total}} = \bigl(1 \times 10^{-3}\ \text{W}\bigr)\,\bigl(0.1\ \text{s}\bigr) = 1 \times 10^{-4}\ \text{J}.$$

Next, we calculate the energy carried by a single photon. The well-known Einstein-Planck relation gives the energy $$E_{\text{photon}}$$ of one photon as

$$E_{\text{photon}} = h \, \nu = \frac{h\,c}{\lambda},$$

where $$h = 6.63 \times 10^{-34}\ \text{J\,s}$$ is Planck’s constant, $$c = 3.00 \times 10^{8}\ \text{m\,s}^{-1}$$ is the speed of light in vacuum, and $$\lambda$$ is the wavelength. The given wavelength is $$\lambda = 1000\ \text{nm}$$. First we express this wavelength in metres:

$$1000\ \text{nm} = 1000 \times 10^{-9}\ \text{m} = 1 \times 10^{-6}\ \text{m}.$$

Now we substitute all values into the photon-energy formula:

$$E_{\text{photon}} = \frac{\bigl(6.63 \times 10^{-34}\ \text{J\,s}\bigr)\,\bigl(3.00 \times 10^{8}\ \text{m\,s}^{-1}\bigr)}{1 \times 10^{-6}\ \text{m}}.$$

Multiplying the numerators and then handling the powers of ten carefully, we obtain

$$E_{\text{photon}} = \frac{6.63 \times 3.00 \times 10^{-34 + 8}}{1 \times 10^{-6}}\ \text{J} = \frac{19.89 \times 10^{-26}}{10^{-6}}\ \text{J}.$$

Dividing by $$10^{-6}$$ is equivalent to multiplying by $$10^{6}$$, so

$$E_{\text{photon}} = 19.89 \times 10^{-26 + 6}\ \text{J} = 19.89 \times 10^{-20}\ \text{J}.$$

For convenience we rewrite this in standard scientific notation by keeping only one non-zero digit before the decimal:

$$E_{\text{photon}} = 1.989 \times 10^{-19}\ \text{J}.$$

Now, the number of photons $$N$$ emitted is the ratio of the total energy radiated to the energy of one photon:

$$N = \frac{E_{\text{total}}}{E_{\text{photon}}} = \frac{1 \times 10^{-4}\ \text{J}}{1.989 \times 10^{-19}\ \text{J}}.$$

Carrying out the division step by step, we first divide the numerical coefficients:

$$\frac{1}{1.989} \approx 0.503.$$

Next we subtract the exponents of ten: $$10^{-4} / 10^{-19} = 10^{(-4) - (-19)} = 10^{15}.$$ Combining these two parts, we get

$$N \approx 0.503 \times 10^{15} = 5.03 \times 10^{14}.$$

The question asks us to express this answer in the form $$x \times 10^{13}$$. Notice that

$$5.03 \times 10^{14} = 50.3 \times 10^{13}.$$

Rounding $$50.3$$ to the nearest integer gives $$50$$. Therefore,

$$x = 50.$$

So, the answer is $$50$$.