When two monochromatic lights of frequency, ν and ν 2 are incident on a photoelectric metal, their stopping potential becomes V S 2 and V s respectively. The threshold frequency for this metal is:

Step-by-step solution

Using Einstein's photoelectric equation we can write, For first case: h ν = ϕ + e V S 2       . . . 1 For second case: h ν 2 = ϕ + e V S       . . . 2 Using value of e V S from equation(2) in equation(1), we can write, h ν = ϕ + 1 2 h ν 2 - ϕ ⇒ h ν = ϕ + h ν 4 - ϕ 2 ⇒ ϕ 2 = 3 h ν 4 ⇒ ϕ = 3 h ν 2 As we know, ϕ = h ν 0 Therefore, ν 0 = 3 ν 2