When photon of energy 4 . 0 eV strikes the surface of a metal A , the elected photoelectrons have maximum kinetic energy T A eV and de-Broglie wavelength λ A . The maximum kinetic energy of photoelectron liberated from another metal B by photon of energy 4 . 50 eV is T B = T A - 1 . 50 eV . If the de-Broglie wavelength of these photoelectrons λ B = 2 λ A , then choose the correct statement(s).

Step-by-step solution

From Einstein photoelectric equation, E = ϕ 0 + KE max For metal A    4 = ϕ A + T A . . . . i For metal B 4 . 5 = ϕ B + T A - 1 . 5 . . . . ii From Eq. (i) and (ii), wo got ϕ B - ϕ A = 2 Now, according to de-Broglie hypothesis, λ A = h m v = h 2 m T A Similarly, λ B = h 2 m T B ∴   λ A λ B = T B T A = T A - 1 . 5 T A = 1 - 1 . 5 T A 1 / 2 1 2 2 = 1 - 1 . 5 T A On solving, T A = 2 . 0 eV or ϕ A = 4 - T A = 4 - 2 = 2 . 0 eV ϕ B = 6 - T A = 6