Question

A sample of hydrogen atoms is irradiated with light with a wavelength of 89.0 nm, and electrons are observed leaving the gas. If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons? A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part A. How can this be?

Answer 1

a)

using the equation

hc/ $\lambda$ = KE_max + W_o

W_o = ground state energy = 13.6 eV

(6.63 x 10^-34) (3 x 10⁸)/(89 x 10^-9) (1/(1.6 x 10^-19)) = KE_max + 13.6

KE_max = 0.37 eV

b)

as the value of "n" increases , the value of W will decrease. hence the kinetic energy of electrons will be greater than the maximum kE in that case