What is the work function of the metal if the light of wavelength 4000 mathring A generates photoelectrons of velocity 6 × 105 ms-1 form it ? (Mass of electron = 9 × 10-31 kg Velocity of light = 3 × 108 ms-1 Planck's constant = 6.626 × 10-34 Js Charge of electron = 1.6 × 10-19 JeV-1)

Question

What is the work function of the metal if the light of wavelength 4000 mathring A generates photoelectrons of velocity 6 × 105 ms-1 form it ? (Mass of electron = 9 × 10-31 kg Velocity of light = 3 × 108 ms-1 Planck's constant = 6.626 × 10-34 Js Charge of electron = 1.6 × 10-19 JeV-1) (A) 0.9 eV (B) 4.0 eV (C) 2.1 eV (D) 3.1 eV

Tardigrade Admin · Accepted Answer

hv =φ + hv° (1/2)mv2 = hc ((1/λ) - (1/λ0)) hv =φ + (1/2) mv2 φ = (6.626 ×10-34×3 ×108/4000×10-10) - (1/2) ×9 ×10-31 ×(6×105)2 φ = 3.35 ×10-19 J ⇒ φ ≃ 2.1 eV

Q. What is the work function of the metal if the light of wavelength 4000A˚ generates photoelectrons of velocity 6×105ms−1 form it ? (Mass of electron = 9×10−31kg Velocity of light = 3×108ms−1 Planck's constant = 6.626×10−34Js Charge of electron = 1.6×10−19JeV−1)

0.9 eV

4.0 eV

2.1 eV

3.1 eV

hv=ϕ+hv∘ 21​mv2=hc(λ1​−λ0​1​) hv=ϕ+21​mv2 ϕ=4000×10−106.626×10−34×3×108​−21​×9×10−31×(6×105)2 ϕ=3.35×10−19J⇒ϕ≃2.1eV