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Asked by Hhhhgffghj Last Modified
Answer It is given that the work function (W0) for caesium atom is 1.9 eV.
(a) From the expression,
, we get:
Where,
λ0 = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of (λ0):
6.53 × 10–7 m
Hence, the threshold wavelength
is 653 nm.
(b) From the expression,
, we get:
Where,
ν0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of ν0:
(1 eV = 1.602 × 10–19 J)
ν0 = 4.593 × 1014 s–1
Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014 s–1.
(c) According to the question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ν0)
= 9.3149 × 10–20 J
Kinetic energy of the ejected photoelectron = 9.3149 × 10–20J
Since K.E 
v = 4.52 × 105 ms–1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105 ms–1.
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