Question 51

The work function for caesium atom is 1.9 eV. Calculate

(a) the threshold wavelength and

(b) 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

A) Work function of caesium (WO) = hvo

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

= 4.59x1014/sec

 

B) λo =c/vo = 3x10/ 4.59x1014 = 6.54x10-7m

 

C) 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

v = 4.53x105ms-1

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