Explanation:
Given that,
Wavelength of the light, [tex]\lambda=4170\ A=4170\times 10^{-10}\ m[/tex]
Work function of sodium, [tex]W_o=4.41\times 10^{-19}\ J[/tex]
The kinetic energy of the ejected electron in terms of work function is given by :
[tex]KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J[/tex]
The formula of kinetic energy is given by :
[tex]KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s[/tex]
Hence, this is the required solution.
