GIVING MEDALS AND FAN 1. Which of the following is the correct statement? A photon of yellow light has more energy than a photon of ultraviolet light. A photon of yellow light has less energy than a photon of ultraviolet light. A photon of yellow light has less energy than a photon of red light. A photon of blue light has more energy than a photon of ultraviolet light. 2. According to classical physics, which of the following should determine the velocity of the photo electrons ejected from a metallic surface? photon energy wavelength frequency intensity
(1) E (energy) = h (Plank's constant) × f (frequency) Which color has the highest frequency?
violet @jcoury
A photon of yellow light has less energy than a photon of ultraviolet light for #1, but what did u get for #2
hiunt: the kinetic energy KE of a photo-electron, is given by the subsequent formula: \[\Large KE = h\nu - W\] where \nu is the frequency of the incident light, and W is the so called "work function" of the material
??
If you look at this link http://www.cyberphysics.co.uk/topics/atomic/Photoelectric%20effect/Photoelectric%20effect.htm you can see in the description on the page that it is photonic energy
W is the work function of the material from which the photo-electrons are emitted
is it wavelength?
more explanation: only the electrons which compose the surface of the metal can be emitted. In order to do that, the incident light have to communicate to those electrons the energy W so those electrons can overcome the potential barrier of the at the surface of that metal
can u tell me the answer please youre just confusing me more
your answer is the fundamental concept in modern physics, since the emitted electrons doesn't depend on the intensity of the incident light, as many phisicist believed, infact the number of emitted electrons depends on the frequency of the incident light, that is another result of the famous scientist Albert Einstein
in other words, the kinetic energy depends on the photon energy, as you can see from my formula above
so the answer is photon energy?
yes! \[\Large \begin{gathered} \frac{1}{2}{m_e}v_e^2 = h\nu - W \hfill \\ \hfill \\ {v_e} = \sqrt {\frac{{2\left( {h\nu - W} \right)}}{{{m_e}}}} \hfill \\ \end{gathered} \]
better is to say frequency, since when we perform an experiment about the photoelectic effect, we consider the frequency of the light used
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