The photoelectric threshold wavelength of silver is $3250 \times 10^{- 10} m$ . The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{- 10} m$ is:-
$(G i v e n h = 4.14 \times 10^{- 15} eV s a n d c = 3 \times 10^{8} m s^{- 1})$

Question

The photoelectric threshold wavelength of silver is

3250 \times 10^{- 10} m

. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength

2536 \times 10^{- 10} m

is:-

(G i v e n h = 4.14 \times 10^{- 15} eV s a n d c = 3 \times 10^{8} m s^{- 1})

Aicharya · Accepted Answer

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

The photoelectric threshold wavelength of silver is 3250 × 10 – 10 m . The velocity of the electron ejected…

Solution