How Close Does the Alpha Particle Get to the Platinum Nucleus?

AI Thread Summary
The discussion revolves around calculating how close an alpha particle can get to a platinum nucleus during a Rutherford-type scattering experiment. The alpha particle has a kinetic energy of 4.0 MeV, and the user attempts to apply energy conservation principles to derive a formula for the distance from the nucleus. There is a clarification that K in the formula refers to Coulomb's constant (9 x 10^9) rather than Planck's constant, which is denoted by h. The user is reassured that they are on the right track, with guidance on correctly identifying the charges involved in the calculation. The conversation emphasizes the importance of understanding the constants and their correct application in the context of the problem.
BubbaJonze
Messages
2
Reaction score
0

Homework Statement



An Alpha particle of a Rutherford-type scattering experiment is shot directly toward a platinum nucleus. One of the particles was apparently aimed directly at the centre of the nucleus cause it shot 180 degrees back along the same path. How close does it come to the platinum nucleus?

Homework Equations



Kinetic energy of Alpha particle: 4.0 MeV

The Attempt at a Solution


Doubt this is right, but I figured energy conservation had something to do with it.
\sumE_{}initial=\sumE_{}final

After some re-arranging, this is what I came up with:
\frac{Kq_{1}q_{2}}{r}=E

So I figured I was solving for r, in which case i got this formula \frac{Kq_{1}q_{2}}{E}
K= Planck's constant
q= elementary charge
E= 4.0MeV (?)

Can someone tell me if I'm anywhere near the right track? We didn't really learn about it, but my instructor stated it was a "challenger" question.
 
Physics news on Phys.org
You are on the right track.

K is not Planck's constant though. K =9X10^9.

Planck's constant, on the other hand, is usually denoted by h and is 6.626X10^-34

Just remember that while q_1=electronic charge, q_2 will be the charge of the other body in the problem, the platinum nucleus.
 
G01 said:
You are on the right track.

K is not Planck's constant though. K =9X10^9.

Planck's constant, on the other hand, is usually denoted by h and is 6.626X10^-34

Just remember that while q_1=electronic charge, q_2 will be the charge of the other body in the problem, the platinum nucleus.
Oh, that's right. I guess I got a little confused since Coulomb's Law was in my last unit and now my constants are all messed up. Thanks.
 
Anytime.:smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top