Light of wavelength λA and λB falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons in KA and KB respectively, then which one of the following relations is true ? (λA = 2 λB)

Question

Light of wavelength λA and λB falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons in KA and KB respectively, then which one of the following relations is true ? (λA = 2 λB) (A) KA < (KB/2) (B) 2 KA = KB (C) KA =2 KB (D) KA > 2 KB

Tardigrade Admin · Accepted Answer

According to Einstein photoelectric equation KA=(h c/λA)-φ ⇒ 2 KA=(2 h c/λA)-2 φ or 2 KA+φ=(2 h c/λA)-φ Wavelength of metal B is half of metal A ∴ 2 KA+φ =KB ⇒ KA <(KB/2)

Q. Light of wavelength $λ_{A}$ and $λ_{B}$ falls on two identical metal plates $A$ and $B$ respectively. The maximum kinetic energy of photoelectrons in $K_{A}$ and $K_{B}$ respectively, then which one of the following relations is true ? $(λ_{A} = 2 λ_{B})$

$K_{A} < \frac{K _{B}}{2}$

$2 K_{A} = K_{B}$

$K_{A} = 2 K_{B}$

$K_{A} > 2 K_{B}$

According to Einstein photoelectric equation
$K_{A} = \frac{h c}{λ _{A}} - ϕ \Rightarrow 2 K_{A} = \frac{2 h c}{λ _{A}} - 2 ϕ$
or $2 K_{A} + ϕ = \frac{2 h c}{λ _{A}} - ϕ$
Wavelength of metal $B$ is half of metal $A$
$∴ 2 K_{A} + ϕ = K_{B}$
$\Rightarrow K_{A} < \frac{K _{B}}{2}$

Q. Light of wavelength λA​ and λB​ falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons in KA​ and KB​ respectively, then which one of the following relations is true ? (λA​=2λB​)

KA​<2KB​​

2KA​=KB​

KA​=2KB​

KA​>2KB​