JEE Main 2026 — Structure of Atom Question with Solution

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 $E=\frac{hc}{\lambda }$:

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

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