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#992989 Photon giving energy ?

Posted by Sensei on Yesterday, 09:12 AM

Are these calculations corrrect ?


Yes. My app is showing 7.99485 MeV as well.

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#992894 Photon giving energy ?

Posted by Sensei on 27 May 2017 - 04:58 PM

So for example: a Helium-4 isotope to a Hydrogen-3 isotope taking 1 proton away would take if my calculations are correct:

Helium-4 has 2 protons and 2 neutrons,

2 * 1,007276466812 = 2,014552934 u    and 2 * 1,00866491600 = 2,017329832

2,014552934 + 2,017329832 = 4,031882766 u

Helium - 4 mass = 4,002603

4,031882766 - 4,002603 = 0,029279766 u

1 u = 931,494061 MeV

0,02927976 * 931,494061 = 27,27392814 MeV 

27,27392814 / 4 = 6,818482034 MeV 

So does that mean that if I have calculated it correctly it takes  6,818482034 MeV to remove a proton of the nucleus of a Helium-4 isotope ?


Not really.


^4_2He + 19.82 MeV \rightarrow ^3_1H + p^+


Take mass of Helium-4 4.0026 u

multiply by 931.494 MeV/u

subtract 2 electrons 0.511 MeV each,

you will receive 3727.38 MeV


Repeat the same with Hydrogen-3  3.01605 u

you will receive 2808.92 MeV.


3727.38 MeV + 19.82 MeV = 2808.92 MeV + 938.272 MeV

(without taking into account momentum)


Helium-4 has one of the largest energies needed to eject proton or neutron, from the all isotopes.

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#992693 Gauging the scale of impact on large prime factorization

Posted by Sensei on 26 May 2017 - 01:11 PM

What David proposed in post #2 is not "Sieve of Eratosthenes".
Sieve of Eratosthenes pseudocode is
 Input: an integer n > 1.
 Let A be an array of Boolean values, indexed by integers 2 to n,
 initially all set to true.
 for i = 2, 3, 4, ..., not exceeding √n:
   if A[i] is true:
     for j = i2, i2+i, i2+2i, i2+3i, ..., not exceeding n:
       A[j] := false.
 Output: all i such that A[i] is true.
(it's begging for using bitwise logic operations and 8 times decrease memory needed for array)
What he proposed in post #2 was something like:
list.append( 2 );
for( int i = 3; i < n; i++ )
  bool found = false;
  for( int j = 0; j < list.count; j++ )
    if( ( i % list[ j ] ) == 0 ) // modulo
     found = true; // divisible by some prime
  if( found == false ) list.append( i ); // append yet another prime to array..
Output array contains nothing but primes.

But better would be to skip the all even numbers, since the beginning:
list.append( 2 );
for( int i = 3; i < n; i += 2 )
  bool found = false;
  for( int j = 1; j < list.count; j++ )
    if( ( i % list[ j ] ) == 0 ) // modulo
     found = true; // divisible by some prime
  if( found == false ) list.append( i ); // append yet another prime to array..

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#992347 What do you think of the idea of a science-themed theme park?

Posted by Sensei on 24 May 2017 - 08:04 PM

It would help if you knew enough English to understand what he was trying to say.

It would help if you knew enough marketing (and how internet search engines work, and how they rank links) to understand when people want to put you something behind the story..

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#992344 What do you think of the idea of a science-themed theme park?

Posted by Sensei on 24 May 2017 - 07:46 PM

That is a ridiculous assertion.

It was clearly an example to show something akin to his idea.

This link point to creationist's ark!
Are you blind?!
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#992340 What do you think of the idea of a science-themed theme park?

Posted by Sensei on 24 May 2017 - 07:35 PM

Damn, if only I had the dough these creationists have. (rubbish link)

Are you here just for advertise this rubbish... ??
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#991672 Personal theories

Posted by Sensei on 21 May 2017 - 03:10 PM

Why is the developer of a scientific theory not allowed to discuss their own theory on this website?

Hijacking means introduction of 3rd party idea/hypothesis/theory into somebody else thread.

If you stay in your own thread discussing there your hypothesis/theory, there will be no hijacking.
At the worst case, thread could be locked by mods, if you won't be participating in discussion (soapboxing), by not answering other members questions and raising doubts about theory or providing counter-arguments disproving it (typically results of experiments rebutting hypothesis).

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#991649 so many questions, so few answers.gg's

Posted by Sensei on 21 May 2017 - 12:28 PM

6-is there anyway we can add or remove the protons and other particles of an atom? i read that the only thing that makes a substance different from others is the amount of particles a nucleus has orbiting it. if so, then why cant we simply change one substance into another?

Yes, we can change one element to other element.
Usually it's done by bombarding it by free neutrons (which have to be created first).
Free neutron has no charge, therefor there is no Coulomb barrier to fight against.
Nuclear reactors are creating free neutrons that can be used in transformation.

But the main problem is scale.

If you have 1 gram of Gold, it has 1 g / 197 g/mol = 0.005 mol
1 mol is 6.022141*10^23 atoms.
Therefor 0.005 mol is 3.06*10^21 Gold atoms..

It's billion multiplied by billion multiplied by 3060.

If you would make billion transformations of one element to other element per second, you would have to wait approximately 100 thousand years to get 1 gram of Gold..

for instance, lead into gold, if you manage to somehow add a few more protons to it, then it should turn to gold. so isnt there a way to to make a nucleus attract the protons it needs to be turned into gold?

Lead has more protons than Gold.
Lead has 82 protons,
Gold has 79 protons.
After capturing proton it would transform to Bismuth, not to Gold.

or maybe even force protons to merge with the atom.

It's professionally called proton capture.

Equivalent for neutrons, neutron capture.

would electromagnetic wave frequencies be able to effect the behavior of a atom or its particles?

Gamma photon with very high energy can destroy atom.
It's called photodisintegration.

Lower energy x-ray and UV photon can ionize atom (electron is ejected from it).

Even lower energy photon can make photoelectric effect in some metals.
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#990965 Relativity Question - Split from Massless Particle

Posted by Sensei on 18 May 2017 - 05:11 AM

Is this photoelectric effect what makes the famous solar panels make electricity?

It could be used.

Is this read "Electron plus Positron yield Photon plus Photon plus 1.022 Million Electron Volts""

Electron and positron (antimatter antiparticle of electron), have the same rest-mass ~ 0.511 MeV/c^2
So, once they annihilate together, they release two photons with energy ~ 0.511 MeV each usually.
Energy prior annihilation is equal to energy post annihilation.
Charge prior annihilation (-1e + 1e) is equal to charge post annihilation (0e).

I do not at all understand the other equation.  I take it is about the momentum of a Photon powering an electron.

Photon energy and momentum is transferred to metal (W), and electron (K.E. of photoelectron). Photon disappears.

If hf<W nothing happens. No photoelectric effect. No photoelectrons ejected from metal.
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#990746 Relativity Question - Split from Massless Particle

Posted by Sensei on 17 May 2017 - 05:38 AM

Please tell me (in words of one syllable) about the experiment involving releasing and detecting a single Photon.



In photoelectric effect, single photon is absorbed, and single electron, called photoelectron, is ejected from metal.


Kinetic energy of photoelectron corresponds to photon energy minus energy required to eject electron.

\frac{1}{2} m_e v^2 = h f - W


When electron and positron annihilates together, there are created typically two photons, going in opposite directions in CoM FoR of matter-antimatter prior annihilation.

e^- + e^+ \rightarrow \gamma + \gamma + 1.022 MeV

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#988820 Chemical identifying equipment

Posted by Sensei on 7 May 2017 - 03:30 PM

I was speaking to a professor at a university and we were discussing identifying different chemicals. He said there is equipment to identify different chemicals. He said that every substance in the universe has a different number and that you can get equipment which will identify each chemical and give you the number for it.

What is the name of this equipment?

It sounds to me that he was talking about mass-spectrometer to layman
But he overestimated it ("every substance in the universe"), or you misunderstood/misinterpret what he said.
Mass-spectrometer can tell you what is mass of compound, and what elements are constituents of compound.
f.e. if you place Gold in mass-spectrometer it will tell you whether it's 14, 18 or 24 carat Gold, and tell you what metal is secondary in alloy.
But couple different compounds can have the same mass, and the same formula. f.e. glucose and fructose have the same chemical formula C6H12O6
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#988713 Turning lead into gold

Posted by Sensei on 6 May 2017 - 06:24 PM

A modern-day Alchemist suggested (speculated!) in these Speculations that we could turn lead into gold by putting a lead atom next to a hydrogen atom (that shouldn't be too hard) and then synchronizing their vibrations until they merge.


"Modern alchemist" should first learn about atomic number of elements.. learn about protonsand neutrons, how they transform, how they decay..



Maybe he's got something there.  Pb + H > Au


Lead has 82 protons in nucleus,
while Gold has 79 protons in nucleus.
If you will merge Pb+H you will get Bismuth (with Z=83 protons), not Gold.


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#988604 Math of relativistic mass different from that of rest mass ?

Posted by Sensei on 6 May 2017 - 06:15 AM

If we start with a steel ball with the same diameter as the Earth the two spacetime regions will effectively be identical.



If you start with steel ball with the same diameter as the Earth, they will have the same volume, but it'll be 43% more massive.

Density of Iron is ~ 7874 kg/m^3, and density of Earth is ~ 5514 kg/m^3

( 7874 kg / m^3 * volume of Earth ) / ( 5514 kg/m^3 * volume of Earth ) = 1.428

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#986881 ionization of gases

Posted by Sensei on 29 April 2017 - 08:31 PM

It mentions CERN used this to contain antiprotons, I thought these were just theoretical, should that have read positrons, or have CERN managed to create an antiproton.


Antiprotons are created since 1955, as you can see on wikipedia:


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#985901 watts and power, converting dw/dt to an integral, and changes in integral int...

Posted by Sensei on 25 April 2017 - 08:56 PM

I am reading Schaum's Outlines, Electric Circuits Fifth Edition having taken both precalculus courses but not calculus.
Firstly, I am confused by the definition of power as the time derivative of watts (p = dw/dt) even though watts is a unit of power.  More importantly, I am confused about how this derivative is reversed into an integral of power as below:
w = \int_0^t p dt

In this context, w is shortcut from "work" rather than "watt".
Work is energy with Joules as unit. Watt = Joules / second.


There is very similar equation in Wikipedia article

 w = \int_0^t p(t) dt


in "Peak power and duty cycle" section.

It could be used to calculate energy of AC pulses.

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