Question

What is the maximum kinetic energy of electrons ejected from barium when illuminated by white light, to 750 nm?

Answer 1

Answer:

$1.0045\times10^{-19}$ 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 ?

$\lambda$ = wavelength of the white light = 400 nm = 400 x 10⁻⁹ m

$c$ = speed of white light = 3 x 10⁸ ms⁻¹

Energy of the white light is given as

$E = \frac{hc}{\lambda} = \frac{(6.63\times10^{-34})(3\times10^{8})}{400\times10^{-9}}\\E = 4.9725\times10^{-19} J$

$W_{o}$ = Work function of barium = 2.48 eV = 2.48 x 1.6 x 10⁻¹⁹ J

$K$ = Maximum kinetic energy

Using photoelectric equation we have

$E = W_{o} + K\\4.9725\times10^{-19} = 2.48\times1.6\times10^{-19} + K\\K = 1.0045\times10^{-19} J$