Question

The work function for caesium atom is 1.9 eV. Calculate the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer 1

Let the work function of caesium (WO) be hvo

Therefore vo = WO/h = 1.9 x1.602 x10-19 / 6.626x10-34 so,

= 4.59x1014/sec

 

so, λo =c/vo = 3x108 / 4.59x1014 = 6.54x10-7m

 

so, K.E of ejected electron = h(v-vo) = hc(1/λ – 1/ λo)

=(6.626x3x10-26) (1/500x10-9 – 1/654x10-9)

=(6.626x3x10-26) / 10-9(154/500x654)

= 9.36x10-20J

k.E = 1mv2/2 = 9.36x10-20J

=9.1x10-31/2 = 9.36x10-20J

(Or)

v2 = 20.55x1010m2s-2

(Or)

Therefore, v = 4.53x105ms-1