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.E_{max} = h\frac {c}{λ} - ф ----------(2)

by putting values

\K.E_{max} = 6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ \frac {3 ｘ 10⁸ m s⁻¹}{3 ｘ 10⁻⁷ m}  - 4.8 ｘ 10⁻¹⁹ J

\K.E_{max} = 6.626 ｘ 10⁻¹⁹ J  - 4.8 ｘ 10⁻¹⁹ J

\K.E_{max} = 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.E_{max} = \frac {1.82 ｘ 10⁻¹⁹ }{1.6 ｘ 10⁻¹⁹ } eV

\K.E_{max} =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.E_{max} of the electron is equal to the work done on ejecting it from a metal surface, so

\K.E_{max}  = W -------------(1)

Where W = q V  = e V₀   

[ from electrostatic V = \frac {w}{q} and here q = e and V =  V₀}

So equation (1)

\K.E_{max}  = e V₀

or 

V₀  = \frac {K.E_{max}}{e}

by putting values

V₀  = \frac {1.14 eV}{e }

e will cancel with each other So,  

V₀  = 1.14 V -----------------Ans. 3



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