Answer:
[tex]1.0045\times10^{-19}[/tex] J
Explanation:
The complete statement is given as
What is the maximum kinetic energy of electrons ejected from barium (Wo = 2.48 eV) when illuminated by white light, 400 nm to 750 nm ?
[tex]\lambda[/tex] = wavelength of the white light = 400 nm = 400 x 10⁻⁹ m
[tex]c[/tex] = speed of white light = 3 x 10⁸ ms⁻¹
Energy of the white light is given as
[tex]E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{400\times10^{-9}}\\E = 4.9725\times10^{-19} J[/tex]
[tex]W_{o}[/tex] = Work function of barium = 2.48 eV = 2.48 x 1.6 x 10⁻¹⁹ J
[tex]K[/tex] = Maximum kinetic energy
Using photoelectric equation we have
[tex]E = W_{o} + K\\4.9725\times10^{-19} = 2.48\times1.6\times10^{-19} + K\\K = 1.0045\times10^{-19} J[/tex]
