Light of two different frequencies whose photons have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function 0.5eV successively. The ratio of maximum speeds of emitted electrons will be

Question

Light of two different frequencies whose photons have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function 0.5eV successively. The ratio of maximum speeds of emitted electrons will be (A) 1:4 (B) 1:2 (C) 1:1 (D) 1:5

Tardigrade Admin · Accepted Answer

Here work function, φ0=0.5eV According to Einstein's photoelectric equation Maximumenergyofemittedelectron=Incidentphotonenergy-Workfunction &there4; K(max)1=1.eV-0.5eV=0.5eV...(i) and K(max)2=2.5eV-0.5eV=2eV...(ii) On dividing equation (i) by equation (ii) , we get (Kmax1/Kmax2)=(0 ⋅ 5 e V/2 e V)=(1/4) ⇒ ((1/2) m vmax12/(1)2 m vmax22)=(1/4) ⇒ (vmax1/vmax2)=√(1/4)=(1/2)

Q. Light of two different frequencies whose photons have energies $1 e V$ and $2.5 e V$ respectively illuminate a metallic surface whose work function $0.5 e V$ successively. The ratio of maximum speeds of emitted electrons will be

$1 : 4$

$1 : 2$

$1 : 1$

$1 : 5$

Q. Light of two different frequencies whose photons have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function 0.5eV successively. The ratio of maximum speeds of emitted electrons will be

1:4

1:2

1:1

1:5

Q. Light of two different frequencies whose photons have energies $1 e V$ and $2.5 e V$ respectively illuminate a metallic surface whose work function $0.5 e V$ successively. The ratio of maximum speeds of emitted electrons will be

$1 : 4$

$1 : 2$

$1 : 1$

$1 : 5$