A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x \times 1020 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 \underline\hspace2cm . (Nearest integer) (h = 6.626 \times 10 - 34 Js)

Structure of Atom PYQ — JEE Main 2021 · 25 Jul · Shift 1 · Q17 with Solution

View written solutionFree

Correct answer: 2

The source gives $1000\,\text{J}$ in $10\,\text{s}$ .

So, power of the source is

P = \frac{1000}{10} = 100\,\text{J s}^{-1}

Thus, energy incident per second is $100\,\text{J}$ .

Given wavelength,

\lambda = 400\,\text{nm} = 400 \times 10^{-9}\,\text{m}

Energy of one photon is

E = \frac{hc}{\lambda}

Substitute values:

E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}}

E = \frac{19.878 \times 10^{-26}}{4 \times 10^{-7}}

E = 4.97 \times 10^{-19}\,\text{J}

If all the supplied energy is in this monochromatic radiation, then

\text{photons per second} = \frac{100}{4.97 \times 10^{-19}}

= 2.01 \times 10^{20}

Since the wavelength is sufficient for photoelectric emission, and one photon can eject at most one electron,

\text{electrons ejected per second} = 2.01 \times 10^{20}

Given this is written as $x \times 10^{20}$ , we get

x \approx 2

\boxed{x = 2}