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.
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.
It is given that the work function ( W0 ) for caesium atom is 1.9 eV.
(a) From the W0=hcλ0 expression, we get:
λ0=hcW0
Where,
λ0 = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of (λ0 ):
λ0=(6.626×10−34Js)(3.0×108ms−1)1.9×1.602×10−19Jλ0=6.53×10−7m
Hence, the threshold wavelength λ0 is 653 nm. (b) From the expression, W0=hv0, we get:
v0=W0h
Where, ν0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of ν0:
v0=1.9×1.602×10−19J6.626×10−34Js(1eV=1.602×10−19J)v0=4.593×1014s−1
Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014s–1.
(c) According to the question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ν0 )
=hc(1λ−1λ0)=(6.626×10−34Js)(3.0×108ms−1)(λ0−λλλ0)=(1.9878×1026Jm)[(653−500)10−9m(653)(500)10−18m2]=(1.9878×10−26J)(153×109)(653)(500)=9.3149×10−20J
Kinetic energy of the ejected photoelectron = 9.3149 × 10–20J
Since K.E =12mv2=9.3149×10−20Jv=√2(9.3149×10−20J)9.10939×10−31kg=√2.0451×1011m2s−2v=4.52×105ms−1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105ms–1.
It is given that the work function ( W0 ) for caesium atom is 1.9 eV.
(a) From the W0=hcλ0 expression, we get:
λ0=hcW0
Where,
λ0 = threshold wavelength
h = Planck’s constant
c = velocity of radiation
Substituting the values in the given expression of (λ0 ):
λ0=(6.626×10−34Js)(3.0×108ms−1)1.9×1.602×10−19Jλ0=6.53×10−7m
Hence, the threshold wavelength λ0 is 653 nm. (b) From the expression, W0=hv0, we get:
v0=W0h
Where, ν0 = threshold frequency
h = Planck’s constant
Substituting the values in the given expression of ν0:
v0=1.9×1.602×10−19J6.626×10−34Js(1eV=1.602×10−19J)v0=4.593×1014s−1
Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014s–1.
(c) According to the question:
Wavelength used in irradiation (λ) = 500 nm
Kinetic energy = h (ν – ν0 )
=hc(1λ−1λ0)=(6.626×10−34Js)(3.0×108ms−1)(λ0−λλλ0)=(1.9878×1026Jm)[(653−500)10−9m(653)(500)10−18m2]=(1.9878×10−26J)(153×109)(653)(500)=9.3149×10−20J
Kinetic energy of the ejected photoelectron = 9.3149 × 10–20J
Since K.E =12mv2=9.3149×10−20Jv=√2(9.3149×10−20J)9.10939×10−31kg=√2.0451×1011m2s−2v=4.52×105ms−1
Hence, the velocity of the ejected photoelectron (v) is 4.52 × 105ms–1.
