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I am a beginner in Physics so please bear with me and keep it as simple as you can.

From what I have learned, "charged particles emit electromagnetic radiation which includes light". So my question is that an object, say a pen, is visible to us even when it is neutral and from what I have learned, it does not emit radiation so we should not be able to see neutral objects such as this pen. But we do see neutral objects so how is this possible?

Thank you and I apologize if this is a "dumb" question.

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  • $\begingroup$ Atoms of an object might be neutral but they are still made of charged subatomic particles and those can emit radiation. $\endgroup$
    – Triatticus
    Commented Aug 5, 2021 at 17:07
  • $\begingroup$ It is the interaction of an object with light from a different source, for example the sun providing sunlight, that you "see". The object itself is not required to directly emit photons. In this way you can see contrast between differently absorbing objects. An emitting object would be visible even in the dark without another light source, but a non emitting wouldn't be visible in that case. $\endgroup$
    – Hans Wurst
    Commented Aug 5, 2021 at 17:10
  • $\begingroup$ @Hans Wurst, correct me if I am wrong but from what I understood is that non-emiting objects which we see "reflect" photons at certain frequency which produces visible light for us to see. Am I right? $\endgroup$
    – Anubis
    Commented Aug 5, 2021 at 19:27
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    $\begingroup$ Imagine a hot piece of metal. It emits its own light despite being neutral. Because, as triatticus points out, neutral materials are still made of charged particles, which can individually interact with and produce photons. $\endgroup$
    – Alwin
    Commented Aug 5, 2021 at 21:40
  • $\begingroup$ @Anubis Yes that happens too. There are a lot of things coming together. Neutral objects also can emit if energy is provided in some form and aren't limited to absorption or scattering. Having a net charge is not required for this. $\endgroup$
    – Hans Wurst
    Commented Aug 6, 2021 at 7:09

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The short answer is that external light bounces off of it to our eyes. If it wasn’t visible due to external light, then we could see it in the dark too.

Most, not all, of the cases where an object is emitting its own light, it’s because of temperature. Loosely, hot things emit light and cold things absorb light. Even light bulbs have some small hot filament or gas or something. There are other, more rare ways to get light from objects too.

When light strikes an object from the outside, it can be transmitted (passes right through), reflected, or absorbed. This is one dynamic for most objects.

The other is the object emitting its own light. If it is cooler than it’s surroundings it will absorb more than emitted. If it is warmer, it will emit more than absorbed. This is unnoticeable for many types of objects and temperature differences, but we can see some stuff glow when hot.

Things that absorb the most light (dark, opaque, not shiny things) are the same ones that emit the most light when hot. The most extreme case is called a black-body by physicists: it doesn’t reflect or pass-through at all. The whole deal is absorb or emit. So if it is cooler than the surroundings, it will be dark black with no light reflecting or being emitted. It will glow the most when hot. (They like it because can isolate and analyze absorb/emit dynamics as a function of temperature). This is how they figured out that the amount of light emitted is a function of temperature to the fourth power.

Emitted light = $k \cdot T^4$, where $k$ is a constant.

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