Ultraviolet light by wavelength 200 nm is incident on the polished surface of Fe (Iron). The work function of the surface is 4.71 eV . What will be its stopping potential? (h = 6 .626 × (10)- 34 J s ; 1 eV = 1 .6 × (10)- 19 J ; C = 3 × (10)8 m s- 1)

Question

Ultraviolet light by wavelength 200 nm is incident on the polished surface of Fe (Iron). The work function of the surface is 4.71 eV . What will be its stopping potential? (h = 6 .626 × (10)- 34 J s ; 1 eV = 1 .6 × (10)- 19 J ; C = 3 × (10)8 m s- 1) (A) 1.5V (B) 2.5V (C) 0.5V (D) None of these

Tardigrade Admin · Accepted Answer

Stopping potential × energy = KEm a x E= φ+kEm a x kEm a x=E- φ =(h C/200 nm)-4.71 eV =6.2 eV-4.71 eV =1.49 eV ∼ 1.5eV Stopping potential V=1.5 V

Q. Ultraviolet light by wavelength $200 nm$ is incident on the polished surface of $F e$ (Iron). The work function of the surface is $4.71 e V$ . What will be its stopping potential?

$(h = 6.626 \times (10)^{- 34} J s; 1 e V = 1.6 \times (10)^{- 19} J; C = 3 \times (10)^{8} m s^{- 1})$

$1.5 V$

$2.5 V$

$0.5 V$

None of these

Stopping potential $\times$ energy = $K E_{ma x}$
$E = ϕ + k E_{ma x}$
$k E_{ma x} = E - ϕ$
$= \frac{h C}{200 nm} - 4.71 e V$
$= 6.2 e V - 4.71 e V$
$= 1.49 e V$ $\sim 1.5 e V$
Stopping potential $V = 1.5 V$

Q. Ultraviolet light by wavelength 200nm is incident on the polished surface of Fe (Iron). The work function of the surface is 4.71eV . What will be its stopping potential? (h=6.626×(10)−34Js;1eV=1.6×(10)−19J;C=3×(10)8ms−1)

1.5V

2.5V

0.5V