Start Free Mock LoginAlready registered? Login here

Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructurePhotoelectric EffectMedium2 minQB

ChemistryMediuminteger

A light source of 32 watt emits light of $λ = 620 nm$ . If all the emitted photons are made to strike a metal plate of work function = 4.5 eV. Then the magnitude of photocurrent is x x $10^{18}$ . [If 40% of the incident photon eject photoelectrons]. Then calculate the value of x.

Answer:

Solution:

To determine if photoemission occurs, compare the photon energy ( $E$ ) with the work function ( $ϕ$ ).

Given $λ = 620$ nm, the photon energy is:

$E = \frac{h c}{λ} = \frac{1240 e Vnm}{620 nm} = 2 e V$

Since $E = 2 e V < ϕ = 4.5 e V$ , no photoelectrons are ejected. Thus, the photocurrent is 0.

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Photoelectric Effect

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...