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

Question

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

Tardigrade Admin · Accepted Answer

From the Einstein's photoelectric equation, KEmax=Eph-φ0 , where Eph,φ0 denotes energy of photon and work function of material. KE1=1eV-0.5eV=0.5eV KE2=2.5eV-0.5eV=2eV (K E1/K E2)=(0 . 5 eV/2 eV)=(1/4) (v1/v2)=√(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 is $0.5 e V$ successively. The ratio of maximum speeds of both 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 is 0.5eV successively. The ratio of maximum speeds of both emitted electrons will be :

1:4

1:2

1:1

1:5

From the Einstein's photoelectric equation, KEmax​=Eph​−ϕ0​ , where Eph​,ϕ0​ denotes energy of photon and work function of material. KE1​=1eV−0.5eV=0.5eV KE2​=2.5eV−0.5eV=2eV KE2​KE1​​=2eV0.5eV​=41​ v2​v1​​=41​​=21​

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 is $0.5 e V$ successively. The ratio of maximum speeds of both emitted electrons will be :

$1 : 4$

$1 : 2$

$1 : 1$

$1 : 5$