Question

The work function of an element is 3.2 eV. Photons of a certain energy are found to eject electrons with a kinetic energy of 6.5 eV.

(a) What is the speed of the ejected electrons?

(c) If the photon energy is increased by a factor of two, what is the kinetic energy of the ejected electrons?

(b) What is the photon frequency?

(Please make numbers legible and with units)

Answer 1

a.) energy = 1/2 x mv² = 6.5 ev

v² = 13 x1.6 x 10^-19/9.1 x 10^-31

v = 1511857 m/s

c.) photon energy, E = work function + KE

Ex2 = work func + new KE.....(i.)

E = work function + initial KE .......(ii.)

multiplying (ii) by 2

2E = 2x (work function) + 2 x initial KE ..........(iii)

substracting i from iii

0 = work function + 2 x initial KE - new KE

=> new KE = work function + 2x initial KE

= 3.2 + 2x 6.5 = 16.2 eV

b.) hv = KE + work func= 6.5+3.2=9.7 e V

=> v = 2.34 x 10¹⁵ Hz