Question

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

Answer 1

`1.0045*10^(-19)` J

Step-by-step 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 = (hc)/(\lambda) = ((6.63*10^(-34))(3*10^(8)))/(400*10^(-9))\\E = 4.9725*10^(-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*10^(-19) = 2.48*1.6*10^(-19) + K\\K = 1.0045*10^(-19) J`