Simple Ideas to Implementation Question (1 Viewer)

[the-derivative]

Hey guys,

I have this simple Ideas to Implementation question, but I can't seem to get the correct answer.

A red laser emitting 600nm wavelength light and a blue laser emitting 450 nm light emit the same power. Compare the rate of emitting electrons.

Thanks guys.

 

[k02033]

assuming that both light are able to eject electrons

comparing how many electrons are emitted per sec is the same as comparing how many photons are emitted from the light source per second. since photoelectric effect is a one to one efffect (one photon ejects one electron)

now power is energy emitted by source per sec. same power for red and blue, let that power be A

now each sec, there are A amount of energy emitted, for red, each photon has energy E=hc/600nm, so in one sec there are A*(600nm/hc) amount of photons released by red source

for blue its A*(450nm/hc) photons per sec. so red source will emit more electrons.

 

[darkchild69]

Hey guys,

I have this simple Ideas to Implementation question, but I can't seem to get the correct answer.

A red laser emitting 600nm wavelength light and a blue laser emitting 450 nm light emit the same power. Compare the rate of emitting electrons.

Thanks guys.

Higher wavelength means lower energy

Power is simply a measurement of energy in a certain time interval.

If they both have the same power, then they both have the same energy emitted in a time interval.

If red light has a lower energy per photon (due to lower frequency), then there will be more photons per unit time travelling past a point.

Assuming that red light was above the threshold frequency, then there will be more photoelectrons emitted, due to an increased number of photons hitting the surface in a certain time interval.

But, however, these photoelectrons will have a lower KE due to the lower energy of the photons of red light.