# Tag Info

1

I think this question does not actually address the photoelectric effect because this would be strongly dependent on the frequency of the light source. To me this seems to be a question regarding the relationship between intensity, power and work. While this is technically not how an electron is emitted in real life, one could calculate the power incident ...

0

For detailed understanding please consult the book "Radiation detection and measurement" by Glenn Knoll. The qualitative answer to your question is, that $\alpha$ particle losses its energy mainly by its interaction with electrons (not nucleus). Hence the electron density is the key factor which decides the range of $\alpha$ particle. if there are N atoms ...

0

Once an air sample is taken, it is possible to analyze the constituent atomic species with high sensitivity using laser cooling in a magneto-optical trap. In particular, the amount of radioactive isotopes can be specified, particularly well for Rubidium and Cesium. I heard this on a conference. Maybe you can find more info if you search in the field of ...

-2

I have been wandering that same question for 40 years...I have asked all kinds of doctors and researchers, only to see a glaze form over their eyes. The question, I think, is not whether it exists, but how do you measure it. The problem is figuring out the machine that can measure it. And then, where to measure, because I think there will be not only an ...

0

The solution you quote actually doesn't conserve momentum. You can use that $p^2$ is a Lorentz invariant and solve $(p_\nu+p_p)^2=(p_e+p_n)^2$, considering the left hand side in the lab frame and the right hand side in the CM frame. You can find $E_\nu$ and check that now momentum is conserved. Anyway, as mentioned in the previous comment, ...

2

I had similar stupid doubt. It's coming from binding energy. The equation you give: Unstable atomic nuclei with an excess of protons may undergo β+ decay $$p → n + e^{+} + \nu_{e}$$ proton decays into neutron, positron, and electron neutrino. How do you think a proton can be converted to neutron which has greater mass? + you get positron and ...

-1

According to wikipedia, it can reach 3500-4000 °C You can increase the temperature by choosing material which is black at visible light range and white at IR range.

6

Your question is identical to a question posed to Feynman by his father. A two-minute monologue on that is here: https://www.youtube.com/watch?v=eebWoZkN3FQ The photons are genuinely created at some moments – and they may be similarly destroyed, too. They're not coming from anywhere and they are going nowhere. The number of photons $N_\gamma$ is simply ...

0

The spectral radiation rate depends on the material in question and the temperature of the material. You can start to learn about this by Googling "Black Body Radiation" and "Planck" . Your terms "glossy black" and "dull white" are far too vague (in a scientific or engineering sense) to be able to answer. Further, the visual color is not necessarily ...

5

For many—but not all—shielding processes the parameters of interest are proportional to the areal density, $$\text{Range} \propto \frac{\text{areal density}}{\text{mass density}} = \frac{\sigma}{\rho} \,,$$ so a first expectation would be that the same areal density of material (of roughly the same composition) will have the same effect. I don't have the ...

1

In addition to the answer by dmckee and to answer the question how high in energy you could get a photon it might be worth thinking about 'Gamma Ray Astronomy' where the highest energy photons are detected. The record highest photon energy observed is apparently currently 80 TeV, which corresponsds to a wavelength of $1.5 \times10^{-20}m$ wavelength (if I ...

4

Higher energy gamma and longer wavelength radio? Keep in mind that the different 'kinds' are merely human labeling conventions for a spectrum that is continuous in the mathematical sense. There is no feature of "radio" that distinguishes it objectively from microwaves. We just pick a boundary on the basis of some technological limitations that apply when ...

2

The microscopic mechanism of emitting photon in a solid is the transformation of kinetic energy of atoms into EM energy. If an atom is in an excited state due to collisions among other atoms, then it will emit photon when it jumps into the ground state, and the energy of the photon is $$E=\varepsilon(\text{excited state})-\varepsilon(\text{ground ... -1 If we did have decay of massless particles then there'd be no reason a photon wouldn't decay from an energy of h\nu into two photons of energy h\nu/2. Eventually all our photons would be red-shifted into oblivion and we'd have no light left in the universe. 0 Yes, the amount of radiation (i.e. power) would decrease N^4 times, since the whole geometry of the system does not change, only the power emitted by the sun. But the temperature of the earth would decrease only N times, since the ratio of the temperatures remains equal - again because of the unchanged geometry. The earth radiates a power proportional ... 1 Background The specific intensity or brightness, I_{\nu}, is defined as:$$ I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1}  where $\nu$ is the frequency, $dE$ the differential energy, $dA$ the differential area, $dt$ the differential time, $d\Omega$ the differential solid angle, and $d\nu$ the differential frequency. We can define a net flux ...

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