Question

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

Answer 1

(D) V + 1.5 volts

Step-by-step 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