On Sep 21, 12:32 pm, "Jeckyl" wrote:
"Y.Porat" wrote in message

ps.com...





On Sep 20, 6:46 pm, "Y.Porat" wrote:
On Sep 20, 3:11 pm, PD wrote:

On Sep 19, 11:39 pm, "Y.Porat" wrote:

On Sep 19, 11:14 pm, PD wrote:

On Sep 19, 10:30 am, "Y.Porat" wrote:

On Sep 19, 3:58 pm, "quentin" wrote:

In Bohr Atom Model, a transition from shell 3 to 2 for hydrogen
atom (Z=1)
results in a photon of wavelength 656 nm:
f=6.564695e-7 -- E=hf = 4.34981e-40 J = 2.7149395e-21 eV

A transition from 4 to 3 results in a photon of wavelength 1875
nm:
f=1.875627e-6 -- E=hf = 1.24280e-39 J = 7.7569700e-21 eV

Now, I'm a little bit perplexed as I would have expected
a smaller photon energy the further away from the nucleus
the electron transition happens. But it seems not to hold.
Can someone verify this pls. Thx.

----------------
and there i s another 'strange thing'

His first "strange thing" was the result of a math error, which a
simple numerical literacy check (does this number make sense?)
quickly
revealed. It's a very basic skill that physics students are trained
on
so that they won't be so easily distracted by "strange things".

while the Proton is fused to get the Alphas particle

Protons don't fuse to make an alpha particle. What do you think an
alpha particle is?

----------------------
the genius Draper whant s to teach me
nclear physics as well (:-)

how do you get the Alpha particle on the sun??

It's quite a complicated process, but you can find it a number of
places on the web, Porat. For the sake of ascii without superscripts
and subscripts, I'll use a S(Z,A) notation. I'll also not mention, for
brevity, the radiated neutrinos and photons.

Initial steps:
p-p: H(1,1) + H(1,1) -- H(1,2) + e(+)
p-e-p: H(1,1) + H(1,1) + e(-) -- H(1,2) {that's deuterium, fyi}
Then H(1,2) + H(1,1) -- He(2,3) {that's a helium isotope but not
an alpha yet)

Then there are three primary branches for He(2,4) production. Two of
them I'll ignore because they are small, multi-step, and actually
involve decays near the end of
Be(4,8) -- He(2,4) + He(2,4)
or
Li(3,7) + H(1,1) -- He(2,4) + He(2,4)
so it's kind of "back door" production of alphas.

The dominant reaction is
He(2,3) + He(2,3) -- He(2,4) + H(1,1) + H(1,1).

There's a proton fusion and a beta-bar decay to generate deuterium,
then there's a deuterium-proton fusion to make helium-3.
Then there's helium-3 fusion to make alphas and more protons.

But you're right. You start with protons and electrons and you end up
with alphas and more protons and more electrons and more positrons and
neutrinos and gammas.

youi start with net hydrogen or protons
and end up with Alphaparticles
th eAlpha particle is the nuc of the helium Atom
soif you make the start situation
and end situation you get that
the mass of the incoming particle (hydrogen)
anf the outcoming products (helium (Alpha) plus .......

ENERGY RADIATION)
there are inrerim procecesses in which
a proton is becoming a Deutreron etc
and a triton but that does not change the fact that
it started with pure hydrogen

now trh ebottom line is that
matter (mass) lost some of itself to procuce
gamma radiation that is photonsas well
ok?
now we see that th eorriginal mass is missing
and energy is gained in return
sowhat happened here
mass was lost ??

do you have experimental evidence that
MASS I S NOT CONSERVED ??!!

That depends on how you count mass. The mass of the whole system,
counting the radiated photons and neutrinos and so on, hasn't changed.
But beware! Counting the masses of a system does NOT mean just doing
an algebraic sum of the rest masses of the individual particles. The
way to find the mass of a system is to measure the total energy of the
system, then measure the total momentum of the system, and then the
mass is given by m^2 = E^2 - p^2. If you find the mass this way,
you'll find that the number hasn't changed one bit.

If, however, you just add up the rest masses of the particles before
and after this process, then yes, you'll find that that the total mass
defined this way is NOT a conserved quantity.

(exactly As in macrocosm ??

Yes, exactly as in macrocosm. As I've already told you, the mass of a
hydrogen H(1,1) atom is less than the mass of the electron plus the
mass of a proton. If you want that in macroscopic terms, the mass of a
mole of H(1,1) is less than the mass of a mole of electrons plus the
mass of a mole of protons. This is known experimentally.

if you claime that in mIrcrocosm it is different
THE BURDAIN OF PROFE IS ON YOU !!!)

No, it's not. It's in the library. I don't owe you a spoon-feeding.
Find it yourself.

and if yes please bring evidence that mass is not conserved .
untill now it is evidence that
mass **is conserved** even IF changing 'phase'
to be in the photon energy !!
and of course it is acording the
momentous formula

E=mc^2 (there is no Gamma factore there
because the gamma factor does not aply whatsoever
to the photon case !! the phton case is an exception case -
in that sense )

PS
now since i am a proffet

Don't flatter yourself.

i predict that the psychopath moron and Superfart--Jekyle=
Poe --
will jump in
so please bypass him

Y.Porat
------------------------

*they loose mass*!!

Why do you think that's strange? When an electron and a proton
"fuse"
to form a hydrogen atom, the combination loses mass. Been known for
decades. Why would this be different?

how strange !! (:-)

Y.Porat
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--------------
you still evaded th emainissue

we got from the imput elements
other elements and energy!!
while the mass of all particles is less compared to the
incoming mass
while the differnce is found exactly in the
energy that was radiated

it is only
the output mass plus the massin energy
thjat is closing the mass eqaution !!
and th econclusion that yiou are afraied to get is
that

E=mc^ of the net eenrgy
includes exacly the missing mass
of the imput mass!!
you see it cleasrly in the above formual
the m there is not for fun
it is mass
and since there is only one kind of mass
(no ****en relativistic mass)
there fore th e mass in energy is
**rest mass or whatever you call it**

no need to involve momentum for obfuscation
it works nicely without momentum consuderstions
do you know why ??
since there is no relativistic mass
NO MOTION WHATSOEVER CAN CREATE
OR KILL MASS!!
it is simply conserved
and the day you get it
let me know about it (it willatke very long time (:-)

now about my being a proffet
you see jsut above i was right
the psychopat Jekyle (the Superfart supporter
and Superfart 'scientist ')
jumped in
exactly as i predicted (:-)

keep well
Y.Porat
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---------------
since Paul Draper keeps on hiding behind the back
of the Psychopath crook idiot Jekyle

LIAR





here is some spoon feed for parrots :

take the mass *in rest situation* of all the masses
of the *incoming* to the pricesds aprticles
call it say A

take the *mass in rest*!! of all the outcoming from
a nuclesr process
and call it B

make A -B and cal it m1
now take the mass m in mc^2 of *all *the energy that
was releast in that process and call it m2
and
you will find that !!:

m1 = m2 exactly !!

So what? Do you think that is different to what others are saying here?

no need to ****en obfuscations with
momentum bisiness as cheating moron do !!

Noone is cheating (except you with your so-called proof s that photons have
mass)

and if you doubt that i can do
(and did one or two of ) those calculations
and amy be even betetr than PD &CO.
(who talk about spoon feedind (:-) --
just have a look at :

http://www.geocities.com/porat_y/mypage.html

Yeup .. utter crap and nonsense. You call that sane physics .. its just the
ramblings of a deranged mind. I assume that once you were at least a
somewhat intelligent human being .. what a tragedy to see you like that.

(And it looks like you know as much about web page design as you do about
physics .. stuff all .. I'd be embarrassed to provide a link to crap like
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-------------------
olease by pass
the psychopath Jekyle

Y.Porat
---------------------------