When light of wavelength λ illuminates the surface of metal 1, the stopping voltage is v. in terms of v, what will be the stopping voltage if the same wavelength is used to illuminate the surface of metal 2? the work function of metal 1 is 3.6 ev and the work function of metal 2 is 2.1 ev. (a) v – 1.5 volts(b) 2.1 volts – v (c) 3.6 volts – v(d) v + 1.5 volts(e) 1.5 volts – v

(D) V + 1.5 volts

Explanation:

Eq. for stopping potential:

eV₀ = hf - ∅  eq (1)

where

V₀ = stopping potential

h = planks's constant

v = frequency of light

f = c/λ

∅ = work function

stopping potential for metal 1 =V₁= V

stopping potential for metal 1 = V₂

work function for metal 1 = 3.6 eVolts

work function for metal 2 = 2.1 eVolts

substituting values for metal 1 in eq (1)

eV₁ = hf - ∅₁

eV = hc/λ - 3.6 eVolts

V = hc/eλ - 3.6 Volts

hc/eλ =  V + 3.6 Volts   eq(2)

Metal 2 is illuminated with same wavelength so stopping potential V₂ is:

eV₂ = hf - ∅₂

eV₂ = hc/λ - 2.1 eVolts

V₂ =  hc/eλ - 2.1 Volts eq(3)

substituting eq(2) in eq(3)

V₂ =   V + 3.6 Volts - 2.1 Volts

V₂ =   V + 1.5 Volts

temp rises, and expansion

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