The surface of sodium metal is irradiated with radiation of...

The work function of sodium in Joules is:

$\Phi = 2.3 \times 1.6 \times 10^{-19} = 3.68 \times 10^{-19}$ J

The kinetic energy of the ejected electrons is:

$K = 2.8 \times 10^{-20} = 0.28 \times 10^{-19}$ J

Using the photoelectric equation, the energy of the incident photon is:

$E = \Phi + K$

$E = 3.68 \times 10^{-19} + 0.28 \times 10^{-19} = 3.96 \times 10^{-19}$ J

The wavelength of the incident radiation is calculated using $E = \dfrac{hc}{\lambda}$:

$\lambda = \dfrac{hc}{E} = \dfrac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.96 \times 10^{-19}}$

$\lambda = \dfrac{19.8 \times 10^{-26}}{3.96 \times 10^{-19}} = 5 \times 10^{-7}$ m

Converting the wavelength to nanometers:

$\lambda = 500 \times 10^{-9}$ m $= 500$ nm

Given that the wavelength is $x$ nm, we get $x = 500$.

Expressing $x$ in the required format:

$x = 5 \times 10^2$

The value to be filled in the blank is $5$.

Answer: $5$