A metal, whose work function is 3.0 eV, is illuminated by the light of wavelength 3 ｘ 10⁻⁷ m. Calculate the (a) threshold frequency, (b) maximum energy of photo-electrons, and (c) stopping potential (Answer: (a) 0.72 ｘ 10¹⁵ Hz, (b) 1.16 eV (c) 1.16V)



Given data:

Work function = ф =  3.0 eV = 3.0 ｘ 1.6 ｘ 10⁻¹⁹ J = 4.8 ｘ 10⁻¹⁹ J

[ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ]

Wavelength of light = λ =  3 ｘ 10⁻⁷ m

∴ 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) Threshold Frequency = f₀ =  ?

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

(c) Stopping Potential = V₀ =  ?



Solution:


Since Work function ф is defined as 

ф = h f₀

or

f₀ = фh 

by putting values 

f₀ = 4.8ｘ10⁻¹⁹J6.626ｘ10⁻³⁴m²kgs⁻¹

f₀ = 0.724 ｘ 10⁻¹⁵ Hz ------------Ans. 1

This is the minimum (Threshold) frequency of the light at which electrons can be ejected from the metal surface.


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

By using Einstein's photoelectric equation to find the maximum energy of an electron

K.Emax = hf - ф --------------(1)

but Frequency f is unknown, so we have

V = f λ

or 

c = f λ      [ v = c ]  

or

f = cλ 

So, equation (1) becomes

K.Emax = hcλ - ф ----------(2)

by putting values

K.Emax = 6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ 3ｘ10⁸ms⁻¹3ｘ10⁻⁷m  - 4.8 ｘ 10⁻¹⁹ J

K.Emax = 6.626 ｘ 10⁻¹⁹ J  - 4.8 ｘ 10⁻¹⁹ J

K.Emax = 1.82 ｘ 10⁻¹⁹ J 

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

K.Emax = 1.82ｘ10⁻¹⁹1.6ｘ10⁻¹⁹ eV

K.Emax =1.14 eV -------------- Ans. 2

This is the maximum energy by which an electrum move after ejecting from the metal surface.



(c) Stopping Potential = V₀ =  ?

As the K.Emax of the electron is equal to the work done on ejecting it from a metal surface, so

K.Emax  = W -------------(1)

Where W = q V  = e V₀   

[ from electrostatic V = wq and here q = e and V =  V₀}

So equation (1)

K.Emax  = e V₀

or 

V₀  = K.Emaxe

by putting values

V₀  = 1.14eVe

e will cancel with each other So,  

V₀  = 1.14 V -----------------Ans. 3



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