NEET2017PhysicsDual Nature of MatterActual
The photoelectric threshold wavelength of silver is 3250 × 10 – 10   m . The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10 – 10   m is:- G i v e n   h = 4.14 × 10 - 15  eV  s     a n d   c = 3 × 10 8  m  s - 1
Options
- A≈ 6 × 10 6 m s - 1
- B≈ 0.6 × 10 6 m s - 1
- C≈ 61 × 10 3 m s - 1
- D≈ 0.3 × 10 6 m s - 1
Correct answer
B. ≈ 0.6 × 10 6 m s - 1
Step-by-step solution
λ 0 = 3250 × 10 - 10  m λ = 2536 × 10 - 10    m ϕ = h c λ 0 = 1242   e V - n m 325   n m = 3.82   eV h ν = h c λ = 1242   e V - nm 253 . 6   nm = 4 .89   eV By Einstein’s photoelectric equation K m a x = h c λ - h c λ 0 K E max = 4.89 - 3.82  eV = 1.07   eV 1 2 m v 2 = 1.07 × 1.6 × 10 - 19 v =   2 × 1.07 × 1.6 × 10 - 19 9.1 × 10 - 31 v = 0.6 × 10 6  m/s