Explanation:
Given that,
Wavelength of the photon, [tex]\lambda=300\ nm=3\times 10^{-7}\ m[/tex]
Work function of the metal, [tex]\phi=1.13\ eV=1.81\times 10^{-19}\ J[/tex]
We need to find the maximum kinetic energy of the ejected electrons. It can be calculated using Einstein's photoelectric equation as :
[tex]hf=E_k+\phi[/tex]
[tex]E_k=hf-\phi[/tex]
[tex]E_k=h\dfrac{c}{\lambda}-\phi[/tex]
[tex]E_k=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{3\times 10^{-7}}-1.81\times 10^{-19}[/tex]
[tex]E_k=4.82\times 10^{-19}\ J[/tex]
[tex]E_k=3.01\ eV[/tex]
or
[tex]E_k=3\ eV[/tex]
So, the maximum kinetic energy of the ejected electrons is 3 ev. Hence, this is the required solution.
