Distance Problem Find Distance Between Alpha Particles

  • Thread starter Thread starter mustang
  • Start date Start date
AI Thread Summary
To find the distance between two alpha particles with an electrical potential energy of 6.35*10^-19 J, the formula PE_electric = k*q1*q2/r is used. Each alpha particle has a charge of 3.20*10^-19 C, leading to the equation (8.99*10^9)(3.20*10^-19)(3.20*10^-19)/r = 6.35*10^-19. Solving for r yields approximately 1.45*10^-9 m. The initial calculation was incorrect due to misreporting the unit as Coulombs instead of meters and providing excessive precision beyond the significant figures of the given potential energy. Accurate calculations confirm the distance between the alpha particles is approximately 1.45*10^-9 m.
mustang
Messages
169
Reaction score
0
Problem.
Given= k_e=8.98755*10^9 Nm^2/C^2, q_p=1.6021*10^-19 C, and g=9.8 m/s^2.
Two alpha particles (helium nuclei), each consisting of two protons and two neutrons, have an electrical potential energy of 6.35*10^-19 J.
What is the distance between these particles at this time? Answer in m.
Note:I figure out that 1 alpha particle is 3.2*10^-19C. So using the Potential energy of electric formula I squared 3.2*10^-19. Afterwards I divided it by 6.35*10^-19 and multplied that by 8.98755*10^9C to get the answer of 1.45313788*10^-9C. However the answer was incorrect what did I do wrong?
 
Physics news on Phys.org
It would be better to write the formula rather than just reciting what you did.

The formula for potential energy for two charged particles with distance r between them is: -(kq1q2)/r (then anti-derivative of the force function with respect to r).

You are given that k=8.99*10^9, that q1 and q2 are both equal to 3.2*10^-19C (actually, you are not given that, you are given that one proton is 1.6021*10^-19 C and an alpha particle has two protons so: 3.20*10^-19 C- Since the potential is given with only 3 sig. figures, I also rounded off k.)and, finally, the the potential is 6.35*10^-19.

That is, the equation says (8.99*10^9)(3.20*10^-19)(3.20*10^-19)/r=
6.35*10^-19 or (92.1*10^-29)/r= 6.35*10^-19. Then 92.1*10^-29= (6.35*10^-19)r so r= (92.1*10^-29)/(6.35*10^-19)= 14.5*10^-10 or
1.45* 10^-9 m.

There are two things wrong with your answer:
1) Since the potential was given with only 3 significant figures, you answer "1.45313788*10^-9" claims too much accuracy.
2) The problem asked for "distance" and specifically said "answer in m" but your answer is "1.45313788*10^-9C" giving the answer in Coulombs as if it were charge!
 
Formula

The formula for this equation is PE_electric=k_c*q1*q2/r. Also when I asked my teacher he said that he didn't get 1.45*10^-9C. He said that I did my math wrong.
 
Also when I asked my teacher he said that he didn't get 1.45*10^-9C. He said that I did my math wrong.

Well, that was pretty much what I said in (2) wasn't it? Please read my response again.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top