An atomic nucleus initially moving at 420m/s emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to 350m/s. If the alpha particle has a mass of 4.0u and the original nucleus has a mass of 222u, what speed does the alpha particle have when it is emitted.
plzzzzzzzzzz help me and give me the formula to find it.
2006-11-27 09:00:15 · 5 answers · asked by nice_ girl 1 in Science & Mathematics ➔ Physics
The momentum after the emission of alpha must remain the same after the emission. Since the formula is p=mv, if the mass changes, the speen must change accordingly.
Before, p=mv=420*222=93240kg.m/s
After, you have two particles. The sum of their two momentum must still be 93240kg.m/s
The remaining nucleus has a mass of 218 and a momentum of:
p=mv=218*350=76300 kg.m/s
The alpha particle has a momentum of
93240-76300=16940 kg.m/s
So you can calculate its speed with
v=p/m=16940/4=4235 m/s.
IF THE ALPHA HAD BEEN EMITTED IN THE OTHER DIRECTION, you would have needed to do a vector addition.
2006-11-27 09:18:38 · answer #1 · answered by kihela 3 · 3⤊ 0⤋
This is a conservation of momentum problem. The change in the nuclear momentum is 70 x 218 (why?), so the change in the alpha's momentum is the same. Thus, its velocity change is 70 x 218/4, which is added to the initial velocity of 420 m/sec to get the alpha's velocity. The speeds specified are too low for relativity to be a factor; in a real problem, the speeds would be much higher and relativity would have to be taken into account.
2006-11-27 09:11:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Ok. Look at the initial condition (before alpha emmission) and final condition (after).
Initially, you have a 222u (m1) particle moving at 420 m/s.(v1) After you have a 218u (222 - 4) (m2) particle moving at 350m/s (v2).
Take the kinetic energy of the bigger mass before and after and
find the difference.
DeltaK= 1/2(m2v2^2 - m1v1^2)
Where did this energy go? Conservation of energy and momentum say it went to the alpha particle.
Kalpha = DeltaK = 1/2 (malpha valpha^2)
valpha = (2*DeltaK/malpha)^-2
You also can solve this using conservation of momentum.
m1v1=m2v2 + malpha * valpha
therefore
valpha = (m1v1-m2v2)/malpha
Hope that helps. (Hope I got that right, too)
2006-11-27 09:11:57 · answer #3 · answered by Michael E 2 · 1⤊ 0⤋
regrettably the horse tail is now no longer accessible. it kinda made me offended, cuz' i luved all those penguins with the ponytails! that is truly a bummer, yet perchance i'm incorrect...sturdy success looking it!
2016-10-07 21:16:15 · answer #4 · answered by grego 4 · 0⤊ 0⤋
the correct answer is:
Walker, Texas ranger
2006-11-27 09:09:01 · answer #5 · answered by SMART @SS 3 · 2⤊ 4⤋