Respuesta :
Given,
The work function of the metal, W=4.48 eV
The work function is given by the formula,
[tex]W=h\upsilon_0[/tex]Where ν₀ is the threshold frequency of the metal.
But the frequency is related to wavelength as,
[tex]\upsilon=\frac{c}{\lambda}[/tex]Where c is the speed of light and λ is the wavelength.
Thus the work function will be,
[tex]W=\frac{hc}{\lambda_0}[/tex]Where λ₀ is the maximum wavelength needed to eject the electron from the given metal.
On substituting the known values in the above equation,
[tex]\begin{gathered} 4.48\times1.6\times10^{-19}\text{ J}=\frac{6.63\times10^{-34}\text{ Js}\times3\times10^8\text{ m/s}}{\lambda_0} \\ \Rightarrow\lambda_0=\frac{6.63\times10^{-34}\text{ Js}\times3\times10^8\text{ m/s}}{_{}4.48\times1.6\times10^{-19}\text{ J}} \\ =277.48\times10^{-9}\text{ m} \end{gathered}[/tex]Thus the maximum wavelength of the light required is 277.48 nm.
The frequency of the light is,
[tex]\begin{gathered} \upsilon_0=\frac{c}{\lambda_0} \\ =\frac{3\times10^8}{277.48\times10^{-9}} \\ =1.08\times10^{15}\text{ Hz} \end{gathered}[/tex]Thus the frequency of the required light is 1.08×10¹⁵ Hz.
