A photocell shows photoelectric effect when green light shines on it. Which of the following is correct if a green light is replaced by a blue light?
Electron emission would get stopped.
Electrons are emitted with more velocity.
Electrons are emitted with less velocity.
Electrons are emitted with same velocity.
My answer would be D, but I want a second opinion
Electrons are emitted with more velocity.
ε(energy of photon) = Work function + KE
ε(blue)>ε(green) => KE(blue) >KE(green)
=> v(blue) > v(green)
Your answer is correct. When a green light shines on a photocell, the photoelectric effect occurs, causing electrons to be emitted. The color of the light affects the energy of the photons, but not the velocity of the electrons. Therefore, when the green light is replaced by a blue light, the electrons will still be emitted with the same velocity. So, the correct answer is D.
To determine the correct answer, we need to understand the principles behind the photoelectric effect. The photoelectric effect is the phenomenon observed when light of sufficient energy strikes a material surface, causing the ejection of electrons from the material. This effect occurs due to the interaction between photons (particles of light) and electrons in the material.
Now, when a green light shines on a photocell, it means that the photons of the green light have enough energy to dislodge electrons from the material's surface. The photoelectric effect will occur, and electrons will be emitted from the photocell with a certain velocity.
If we replace the green light with a blue light, which has higher energy photons than green light, the photoelectric effect is likely to occur as well. The higher energy photons of the blue light will have enough energy to dislodge electrons from the material. Therefore, electrons will be emitted from the photocell, but now with a higher velocity.
This means that the correct answer is B: Electrons are emitted with more velocity when a blue light is used instead of a green light.