The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Å is 0 . 42 V . If the threshold frequency is x × 10 13 s , where x is (nearest integer): (Given, speed light = 3 × 10 8 m s - 1 . Planck's constant = 6 . 63 × 10 - 34 J s )

Step-by-step solution

We know that, K E m a x = h f - ϕ = h c λ - ϕ Or, h c λ - ϕ = e V 0 , where, ϕ = h ν th and V 0 is stopping potential. So, threshold frequency is ν th = c λ - e V 0 h = 3 × 10 8 66330 × 10 - 10 - 1 . 6 × 10 - 19 × 0 . 42 6 . 63 × 10 - 34 = 3 6 . 630 × 10 15 - 1 . 6 × 0 . 42 6 . 63 × 10 15 = 10 15 3 6 . 630 - 1 . 6 × 0 . 42 6 . 63 = 0 . 4524 - 0 . 1013 ν th = 35 . 11 × 10 13