A source of monochromatic radiation wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, $$x \times 10^{20}$$ electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is ___. (Nearest integer)
h = 6.626 $$\times 10^{-34}$$ Js

We begin by noting the data given in the problem. The source supplies a total energy of 1000 J in 10 s, therefore the energy delivered each second (the power) is

$$P \;=\; \frac{1000\ \text{J}}{10\ \text{s}} \;=\; 100\ \text{J s}^{-1}.$$

This means that every second, 100 J of radiant energy strike the sodium surface.

Next we must find the energy carried by one photon of the monochromatic light. For a photon of wavelength $$\lambda$$ the energy is given by the Planck relation

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

where $$h = 6.626 \times 10^{-34}\ \text{J s}$$ and $$c = 3.00 \times 10^{8}\ \text{m s}^{-1}.$$ The wavelength is

$$\lambda \;=\; 400\ \text{nm} \;=\; 400 \times 10^{-9}\ \text{m}.$$

Substituting these numbers,

$$E_{\text{photon}} = \frac{(6.626 \times 10^{-34}\,\text{J s}) (3.00 \times 10^{8}\,\text{m s}^{-1})} {400 \times 10^{-9}\,\text{m}}.$$

Simplifying step by step, first multiply the numerators:

$$6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 19.878 \times 10^{-26}\ \text{J m}.$$

Rewrite the denominator clearly:

$$400 \times 10^{-9}\ \text{m} = 4.00 \times 10^{2} \times 10^{-9}\ \text{m} = 4.00 \times 10^{-7}\ \text{m}.$$

Now divide:

$$E_{\text{photon}} = \frac{19.878 \times 10^{-26}}{4.00 \times 10^{-7}} = 4.9695 \times 10^{-19}\ \text{J}.$$

Thus, each photon carries approximately

$$E_{\text{photon}} \;\approx\; 4.97 \times 10^{-19}\ \text{J}.$$

The number of photons incident every second is then

$$N_{\text{photon\;per\;sec}} = \frac{\text{energy per second}}{\text{energy per photon}} = \frac{100\ \text{J s}^{-1}}{4.97 \times 10^{-19}\ \text{J}} = 2.01 \times 10^{20}\ \text{photons s}^{-1}.$$

The wavelength is sufficient for photo-emission, and we assume a one-to-one correspondence between photons absorbed and electrons ejected. Therefore

$$N_{\text{electron\;per\;sec}} = N_{\text{photon\;per\;sec}} \approx 2.01 \times 10^{20}\ \text{electrons s}^{-1}.$$

This can be written in the form $$x \times 10^{20}$$ electrons per second, giving

$$x \;\approx\; 2.01.$$

Rounding to the nearest integer, $$x = 2.$$

So, the answer is $$2$$.