## Properties of Light

$c=\lambda v$

Lusin_Yengibaryan_3B
Posts: 103
Joined: Wed Sep 30, 2020 9:33 pm

### Properties of Light

Is c=hv describing the photoelectric effect or some other experiment?

Kimiya Aframian IB
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Joined: Wed Sep 30, 2020 9:34 pm

### Re: Properties of Light

Lusin_Yengibaryan_2A wrote:Is c=hv describing the photoelectric effect or some other experiment?

Hi! I don't think that "c=hv" is an equation because the original would be "E=hv", but "E (energy)" and "c (speed of light)" are not the equivalent (you can see that with their respective units). The relationship between "c" and "v" is that "c=lamda*v"
Hope this helps!

Ria Nawathe 1C
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### Re: Properties of Light

As the previous answer stated, the equation is E = hv. The photoelectric effect demonstrated that light exhibits properties of particles and also demonstrated the relationship between energy and frequency of photons.

Akash J 1J
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### Re: Properties of Light

I don't think c=hv is a valid equivalency. However, I think you may have mixed up the two equivalencies "c = wavelength x freq" and "E = h x freq".

Neither of these equations directly describes the photoelectric effect, but they do create easy ways to convert between units.
c = wavelength x freq is useful for converting between wavelength and frequency and E = h x freq is useful for converting between energy in Joules to frequency in Hz.

Malakai Espinosa 3E
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### Re: Properties of Light

Like all the previous posts said, the equation would be E=hv. This equation does not directly describe the photoelectric effect, but is merely a component in explaining the experiment. E=hv actually describes the energy of a photon, which we can then use along with the work function to show the photoelectric effect.

Lusin_Yengibaryan_3B
Posts: 103
Joined: Wed Sep 30, 2020 9:33 pm

### Re: Properties of Light

Yes, I meant c=lamda * v, but I thought h looked the closest to lamda since I couldn't type lamda on my keyboard.