Yellow light 577nm wavelength is incident on a cesium surface. The stopping voltage is found to be 0.25 V. Find
(a) the maximum K.E. of the photoelectrons
and (b) the work function of cesium. (Answer: (a) 4 ｘ 10⁻²⁰ J (b) 1.9 eV )


Given data:

Wavelength of yellow light = λ =  577 nm = 577 ｘ 10⁻⁹ m
 
Stoppage Voltage = V₀ = 0.25 V


∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹

∴ Velocity of light = c =  3 ｘ 10⁸ m s⁻¹


To Find:

(a) Maximum energy of the photo-electron = K.E =  ?

(b) Work function = ф = ?



Solution:

(a) Maximum energy of the photo-electron = K.E =  ?

The formula for finding the maximum energy of an electron according to given data is

K.Emax =  V₀ e


by putting values

K.Emax =  0.25 V ｘ 1.6 ｘ 10⁻¹⁹ C

K.Emax = 0.4 ｘ 10⁻¹⁹ J

or

K.Emax = 4 ｘ 10⁻²⁰ J -----------Ans.1


This is the maximum energy by Photo-electron.



(b) Work function = ф = ?

Using Einstein's photoelectric equation to find the maximum energy of the electron

K.Emax = hf - ф 

or
 
ф = hf - K.Emax

but Frequency f is unknown, so we have

V = f λ

or 

c = f λ      [ v = c ]  

or

f = cλ 

So, equation (1) becomes

ф = hcλ - K.Emax ----------(2)

by putting values

ф = 6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ 3ｘ10⁸ms⁻¹577ｘ10⁻⁹m - 4 ｘ 10⁻²⁰ J

ф = 3.45 ｘ 10⁻¹⁹ J  - 0.4 ｘ 10⁻¹⁹ J

ф = 3.05 ｘ 10⁻¹⁹ J 

by expressing in electron Volts (eV) we have [ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ] So, using the unitary method we have

K.Emax = 3.05ｘ10⁻¹⁹1.6ｘ10⁻¹⁹ eV

K.Emax =1.9 eV -------------- Ans.2


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