Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1 and λ2, respectively are incident on a metallic surface. If λ1=3 λ2 then:

Question

Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1 and λ2, respectively are incident on a metallic surface. If λ1=3 λ2 then: (A) K 1>( K 2/3) (B) K 1<( K 2/3) (C) K 1=( K 2/3) (D) K 2=( K 1/3)

Tardigrade Admin · Accepted Answer

( hc /λ1)-φ= K 1 ( hc /λ2)-φ= K 2 λ1=3 λ2 3 K 1=(3 hc /λ1)-3 φ 3 K 1=( hc /λ2)-3 φ 3 K 1= K 2-2 φ 3 K 1< K 2 K 1<( K 2/3)

Q. Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $λ_{1}$ and $λ_{2}$ , respectively are incident on a metallic surface. If $λ_{1} = 3 λ_{2}$ then:

$K_{1} > \frac{K _{2}}{3}$

$K_{1} < \frac{K _{2}}{3}$

$K_{1} = \frac{K _{2}}{3}$

$K_{2} = \frac{K _{1}}{3}$

$\frac{h c}{λ _{1}} - ϕ = K_{1}$
$\frac{h c}{λ _{2}} - ϕ = K_{2}$
$λ_{1} = 3 λ_{2}$
$3 K_{1} = \frac{3 h c}{λ _{1}} - 3 ϕ$
$3 K_{1} = \frac{h c}{λ _{2}} - 3 ϕ$
$3 K_{1} = K_{2} - 2 ϕ$
$3 K_{1} < K_{2}$
$K_{1} < \frac{K _{2}}{3}$

Q. Let K1​ and K2​ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1​ and λ2​, respectively are incident on a metallic surface. If λ1​=3λ2​ then:

K1​>3K2​​

K1​<3K2​​

K1​=3K2​​

K2​=3K1​​