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.
Given:
Work Function for Caesium, W = 1.9 eV
To find threshold frequency:
W = hvo
Where
h = Planck’s constant
vo = threshold frequency
vo = h / w
vo = [1.9× 1.602× 10-19] / [6.626× 10-34]
= [3.0438× 10-19] / [6.626× 10-34]
= 4.5937× 1014 Hz
Therefore, the frequency is 4.59× 1014 Hz
To find wavelength:
Speed of Light = [Frequency] × [Wavelength]
We know speed of light = 3×108 m/s
Wavelength, λ = [3×108] / [4.5937× 1014]
= 6.5306× 10-7 m
Therefore, the wavelength is 6.53× 10-7 m
Finding Kinetic Energy:
K.E of ejected electron = h[v-vo] = hc[1/λ – 1/λ o]
= [6.626× 3× 10-26] [1/500× 10-9 – 1/654× 10-9]
= [6.626× 3× 10-26] × {109 × [154/327000]}
= [6.626× 3× 10-26] × [468747.1289]
= 9.3177× 10-20J
Finding Velocity of photoelectron:
We know the formula for kinetic energy which is given as follows:
1/2 mv2 = kinetic Energy
Where
m = mass of electron
v = velocity of electron
v2 = [2× 9.3177× 10-20] / [9.1× 10-31]
= [1.8654× 10–19] / [9.1× 10-31]
v2 = 2.0478× 1011
Therefore v = √ [2.0478× 1011]
v = 4.525× 105 m/s
Therefore, the velocity of photoelectron is 4.525× 105 m/s
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