Light of wavelength 4000 Å is allowed to fall on a metal surface having work function 2 eV . The maximum velocity of the emitted electrons is (.h=6.6× (10)- 34 Js )

Question

Light of wavelength 4000 Å is allowed to fall on a metal surface having work function 2 eV . The maximum velocity of the emitted electrons is (.h=6.6× (10)- 34 Js ) (A) 1.35× 105 m s- 1 (B) 2.7× 105 m s- 1 (C) 6.2× 105 m s- 1 (D) 8.1× 105 m s- 1

Tardigrade Admin · Accepted Answer

(1/2)mv2=(h c/λ )-φ (.in eV.) =(6.6 × 10- 34 × 3 × 108/4000 × 10- 10 × 1.6 × 10- 19)-2 =3.1-2=1.1 eV=1.1× 1.6× 10- 19 J =1.76× 10- 19 J v=(1.76 × 10- 19 × 2/9 × 10-3 1) =6.2× 105 m s- 1

Q. Light of wavelength $4000 Å$ is allowed to fall on a metal surface having work function $2 e V$ . The maximum velocity of the emitted electrons is $(h = 6.6 \times (10)^{- 34} J s$ )

$1.35 \times 1 0^{5} m s^{- 1}$

$2.7 \times 1 0^{5} m s^{- 1}$

$6.2 \times 1 0^{5} m s^{- 1}$

$8.1 \times 1 0^{5} m s^{- 1}$

Q. Light of wavelength 4000Å is allowed to fall on a metal surface having work function 2eV . The maximum velocity of the emitted electrons is (h=6.6×(10)−34Js )

1.35×105ms−1

2.7×105ms−1

6.2×105ms−1

8.1×105ms−1

21​mv2=λhc​−ϕ(ineV) =4000×10−10×1.6×10−196.6×10−34×3×108​−2 =3.1−2=1.1eV=1.1×1.6×10−19J =1.76×10−19J v=9×10−311.76×10−19×2​ =6.2×105ms−1

Q. Light of wavelength $4000 Å$ is allowed to fall on a metal surface having work function $2 e V$ . The maximum velocity of the emitted electrons is $(h = 6.6 \times (10)^{- 34} J s$ )

$1.35 \times 1 0^{5} m s^{- 1}$

$2.7 \times 1 0^{5} m s^{- 1}$

$6.2 \times 1 0^{5} m s^{- 1}$

$8.1 \times 1 0^{5} m s^{- 1}$