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Chemistry Question on Atomic Models

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

It is given that the work function (W0) for caesium atom is 1.9 eV.
(a) From the expression,W0 = hcλ0\frac{hc}{\lambda_0} , we get: λ0 = hcW0\frac{hc}{W_0}
Where, W0 = threshold wavelength
h = Planck's constant
c = velocity of radiation
Substituting the values in the given expression of W0: λ0 = (6.626×10−34Js)(3.0×108ms−1)1.9×1.602×10−19J\frac{(6.626\times10^{-34}Js)(3.0\times10^8ms^{-1})}{1.9\times1.602\times10^{-19}J}
λ0 = 6.53 × 10-7 m
Hence, the threshold wavelength λ0 is 653 nm.

(b) From the expression, W0 = hv0 , we get: v0 = W0h\frac{W_0}{h}
Where, v0= threshold frequency
h = Planck's constant
Substituting the values in the given expression of W0: v0 = 1.9×1.602×10−19J6.626×10−34Js\frac{1.9\times1.602\times10^{-19}J}{6.626\times10^{-34}Js}
(1 eV = 1.602 × 1019 J)
W0= 4.593 × 1014 s-1
Hence, the threshold frequency of radiation (W0) is 4.593 × 10 14 s-1.

(c) According to the question: Wavelength used in irradiation = 500 nm
Kinetic energy = hc (1λ−1λ0\frac{1}{\lambda}-\frac{1}{\lambda_0}) = (6.626×10-34 Js) (3.0×108 ms-1) (λ0 - λ / λλ0) = (1.9878 × 10-26 Jm) [(653−500)10−9m(653)(500)10−18m2\frac{(653 - 500) 10 - 9 m}{(653) (500) 10 - 18 m2}]
=(1.9878×10−26)(153×109)(653)(500)J\frac{(1.9878 × 10^{-26}) (153 × 10^9)}{(653)(500)}J = 9.3149 × 1020 JJ
The kinetic energy of the ejected photoelectron = 9.3149 × 1020J
Since K.E = 12\frac{1}{2}mv2 = 9.3149 × 10-20 J
v = √2(9.3149×10−20J)9.10939×10−31\frac{√2 (9.3149 × 10^{-20J})}{9.10939 × 10^{-31}} kg = √2.0451 × 1011 m2 s-2
v = 4.52 × 105 ms-1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105 ms-1.