When light of wavelength 248   nm falls on a metal of threshold energy 3 . 0   eV , the de-Broglie wavelength of emitted electrons is ________ A o . (Round off to the Nearest Integer). [Use : 3 = 1 . 73 ,   h = 6 . 63 × 10 - 34   Js ;   m e = 9 . 1 × 10 - 31   kg ;   c = 3 . 0 × 10 8   ms - 1 ;   1   eV = 1 . 6 × 10 - 19   J ]

Step-by-step solution

Energy incident = hc λ = 6 . 63 × 10 - 34 × 3 . 0 × 10 8 248 × 10 - 9 × 1 . 6 × 10 - 19   eV = 6 . 63 × 3 × 100 248 × 1 . 6 = 0 . 05   eV × 100 = 5   eV Now using E = ϕ + K . E 5 = 3 + K . E . K . E . = 2 eV = 3 . 2 × 10 - 19   J for de Broglie wavelength λ = h mv K . E . = 1 2 mv 2 so v = 2 K . E . m hence λ = h 2 K . E . × m = 6 . 63 × 10 - 34 2 × 3 . 2 × 10 - 19 × 9 . 1 × 10 - 31 = 6 . 63