A light of wavelength 3000 A o falls on a metal surface. Ejected e - is further accelerated by a potential difference of 2 V , then final K.E of the e - is found to be 8 × 1 0 - 19 J . If threshold energy for the metal surface is ' ϕ ' e V . Then find the numerical value of 8 ϕ

Step-by-step solution

KE final = E photon + E potential − ϕ where ϕ = threshold energy for metal surface E photon = hc λ = 6.6 × 10 − 34 × 3 × 10 8 3000 × 10 − 10 ≃ 6.6 × 10 − 19 J = 4.12 eV E potential = 2 eV KE final = 8 × 10 − 19 J 1.6 × 10 − 19 J = 5 eV Hence, ϕ = ( 4.12 + 2 − 5 )   eV = 1.125   eV Numerical value = 8 × 1.125 = 9