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who knows what recoil velocity is?? (1 Viewer)
- Thread starter crynow
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BlackJack
Vertigo!
wow.... where did that come from? I've met a similar one before... In a moment...
We want the momentum of the photon that came out... which is equal to the momentum of the hydrogen backwards...
l is wavelength.
1/l = R (1^2 - 1/6^2) = 1.0665...*10^7
l=h/mv
mv=h * 1/l
=7.0668...*10^-27
velocity of hydrogen, divide by mass of H... mass=1.673e-27 + 9.109e-31
v= 4.2217.... m/s
Anyone else?
We want the momentum of the photon that came out... which is equal to the momentum of the hydrogen backwards...
l is wavelength.
1/l = R (1^2 - 1/6^2) = 1.0665...*10^7
l=h/mv
mv=h * 1/l
=7.0668...*10^-27
velocity of hydrogen, divide by mass of H... mass=1.673e-27 + 9.109e-31
v= 4.2217.... m/s
Anyone else?
The Spruce Moose
Member
Black Jack 
superhubert
Member
Blackjacks on fire.... It's all about conservation of momentum guy's, (Goddard anyone?) but i thought recoil velocity was usually the name given to the reverse movement of a cannon or whatever when a projectile was lauched.. in this case don't they give it a different name for something like a ejection velocity or just simply the change in momentum of the hydrogen atom or the electron. in any case i was of the asumption change in momentum was limited to rocket thrust in our sylabus. does anyone still actually believe in the sylabus...............:apig:
BlackJack
Vertigo!
My friend's been following the syllabus to the letter, and he topped our shchool in phys this year... I still believe in the syllabus.
I go to Caringbah, and I didn't top anything this year. If lucky, 95% in 3u and 80% in 4u...
I think they used a combination of two dotpoints in this years syllabus, and the conservation of momentum we learnt in the past 9.8.1.(9) with solve problems using rydberg's stuff and 2.(4) with solve problems using the Plank//Einstein stuff...
Those are the two equations I used.
Where did this question come from?
edit: I would like to purport that I'm not a genius, if you want one, go ask spice girl,
I still believe in the syllabus.I go to Caringbah, and I didn't top anything this year
. If lucky, 95% in 3u and 80% in 4u...I think they used a combination of two dotpoints in this years syllabus, and the conservation of momentum we learnt in the past 9.8.1.(9) with solve problems using rydberg's stuff and 2.(4) with solve problems using the Plank//Einstein stuff...
Those are the two equations I used.
Where did this question come from?
edit: I would like to purport that I'm not a genius
, if you want one, go ask spice girl,superhubert
Member
4U....glutton for punishment....
Jellymonsta
Member
q2q i think
superhubert
Member
no i don't think so...i don't do q2q, it's just a combination of space and quantum theory from ideas to implimentation..isn't it?
Jellymonsta
Member
isnt all the Rydberg boojum from Q2Q? i refuse to actually check. im studying for english extension
BlackJack
Vertigo!
I used both from the Q2Q. This question is for Q2Q i'm sure. it's not in ideas suyllabus, but in Q2Q.
BlackJack
Vertigo!
I think they might be able argue that it's in the syllabus... combination of two formulas and a law that we already know?
Hey guys, I tried posting my answers yesturday but I always get fucking database errors - they piss me fucking off. *mummbles*
I did the question two ways, the first of which was the way in which blackjack did it, and I got the same answer. Although I might point out that if you got a question like that, mention that one can't find the velocity (a vector), only the speed.
The other way I did it, which I admit i'm not too sure about, was by equating kinetic energy gained by the photon (13.25 eV) with the kinetic energy gained by the hydrogen atom, which was originally at rest.
i.e originally E (hydrogen atom) = 0 because it was at rest.
but afterwards, shouldn't it equal 13.25 eV? By doing it this way you get an answer of around 50000m/s which is remarkably different from the 4m/s obtained by equating conservation of momentum.
Anyway, I think this last example is somewhat flawed, but the thought did cross my mind.
I did the question two ways, the first of which was the way in which blackjack did it, and I got the same answer. Although I might point out that if you got a question like that, mention that one can't find the velocity (a vector), only the speed.
The other way I did it, which I admit i'm not too sure about, was by equating kinetic energy gained by the photon (13.25 eV) with the kinetic energy gained by the hydrogen atom, which was originally at rest.
i.e originally E (hydrogen atom) = 0 because it was at rest.
but afterwards, shouldn't it equal 13.25 eV? By doing it this way you get an answer of around 50000m/s which is remarkably different from the 4m/s obtained by equating conservation of momentum.
Anyway, I think this last example is somewhat flawed, but the thought did cross my mind.
Jellymonsta
Member
In the Success One HSC Physcs Questions/Answers book:
calculate the momentum of a photon with a wavelength of 388.5nm and the recoil velocity of a (2/1)H atom that emits the photon. (3 marks)
that book was written by the STANSW. so im sure BoS can ask that type of Q.
calculate the momentum of a photon with a wavelength of 388.5nm and the recoil velocity of a (2/1)H atom that emits the photon. (3 marks)
that book was written by the STANSW. so im sure BoS can ask that type of Q.
Oh and I nearly forgot, recoil velocity is just the velocity of a gun,cannon, atom etc once a projectile/particle has been projected/emmited. It is a direct consequence of the conservation of momentum (which further leads me to believe i'm not considering all the factors with my second example above).