A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is $$x \times 10^{20}$$. The value of x is _________. (Nearest integer)
[Given: h = $$6.63 \times 10^{-34}$$ Js and c = $$3.0 \times 10^8$$ ms$$^{-1}$$]

We have a bulb whose power is given as $$P = 50 \text{ W}$$. Power tells us the energy supplied per unit time, that is $$1 \text{ W} = 1 \text{ J s}^{-1}$$, so the bulb emits $$50 \text{ J}$$ of energy every second.

The light is monochromatic with wavelength $$\lambda = 795 \text{ nm}$$. First convert the wavelength into metres:

$$\lambda = 795 \text{ nm} = 795 \times 10^{-9} \text{ m} = 7.95 \times 10^{-7} \text{ m}.$$

The energy of a single photon is obtained from the Planck-Einstein relation, which we state explicitly:

Energy of one photon: $$E = \dfrac{h c}{\lambda},$$ where $$h = 6.63 \times 10^{-34} \text{ J s}$$ is Planck’s constant and $$c = 3.0 \times 10^{8} \text{ m s}^{-1}$$ is the speed of light.

Substituting the given values, we get

$$E = \dfrac{(6.63 \times 10^{-34}) (3.0 \times 10^{8})}{7.95 \times 10^{-7}} = \dfrac{1.989 \times 10^{-25}}{7.95 \times 10^{-7}}.$$

Now divide the mantissas and handle the powers of ten separately:

$$\dfrac{1.989}{7.95} = 0.25,$$ $$10^{-25} \div 10^{-7} = 10^{-25 + 7} = 10^{-18}.$$

So

$$E = 0.25 \times 10^{-18} \text{ J} = 2.5 \times 10^{-19} \text{ J}.$$

Let $$N$$ be the number of photons emitted per second. The total energy emitted in one second is the power $$P$$, so

$$P = N E \quad \Longrightarrow \quad N = \dfrac{P}{E}.$$

Substituting $$P = 50 \text{ J s}^{-1}$$ and $$E = 2.5 \times 10^{-19} \text{ J},$$ we get

$$N = \dfrac{50}{2.5 \times 10^{-19}} = \left(\dfrac{50}{2.5}\right) \times 10^{19} = 20 \times 10^{19} = 2.0 \times 10^{20}.$$

The question writes the result as $$x \times 10^{20}$$, so we compare:

$$N = 2.0 \times 10^{20} \; \Longrightarrow \; x = 2.$$

So, the answer is $$2$$.