- #1
Jimmy87
- 686
- 17
Hi,
Please could somebody verify if what I have understood about the UV catastrophe is along the correct lines. I used this hyperphysics link:
(http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html)
Is it saying that using classical physics an object like a pen should be emitting just as much UV light as infrared? The way I understand it is that an object like a pen is absorbing energy all day but also emitting the energy back and this energy being emitted back should have equal probability of being infrared, visible, UV (as per the table in the above link)? So as an example, let's say after 5s a pen absorbs 100eV of energy from the room. Since UV photons have an energy of around 10eV then this energy could hypothetically (at least classically) be emitted back as 10 UV photons rather than what actually happens which is that 100 infrared photons (1eV each) are emitted back. Is this basically what classical physics had to say about what EM radiation objects should emit?
Could someone also explain why a gold leaf electroscope has to be charged before it demonstrates the photoelectric effect. In all online demos I have seen a charged rod is brought close and the person touches the top to allow electrons to flow onto the electroscope causing the leaf to stay deflected after the rod is removed. Why can't a neutral electroscope be used? If it is not charged initially then the leaf would be down but if you shine UV light on the top then surely electrons would still come off leaving the electroscope positively charged and therefore the leaf would start to rise?
Please could somebody verify if what I have understood about the UV catastrophe is along the correct lines. I used this hyperphysics link:
(http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html)
Is it saying that using classical physics an object like a pen should be emitting just as much UV light as infrared? The way I understand it is that an object like a pen is absorbing energy all day but also emitting the energy back and this energy being emitted back should have equal probability of being infrared, visible, UV (as per the table in the above link)? So as an example, let's say after 5s a pen absorbs 100eV of energy from the room. Since UV photons have an energy of around 10eV then this energy could hypothetically (at least classically) be emitted back as 10 UV photons rather than what actually happens which is that 100 infrared photons (1eV each) are emitted back. Is this basically what classical physics had to say about what EM radiation objects should emit?
Could someone also explain why a gold leaf electroscope has to be charged before it demonstrates the photoelectric effect. In all online demos I have seen a charged rod is brought close and the person touches the top to allow electrons to flow onto the electroscope causing the leaf to stay deflected after the rod is removed. Why can't a neutral electroscope be used? If it is not charged initially then the leaf would be down but if you shine UV light on the top then surely electrons would still come off leaving the electroscope positively charged and therefore the leaf would start to rise?