Help With Finding The Speed of an Alpha Particle

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The discussion focuses on calculating the speed of an alpha particle emitted during the alpha decay of a uranium nucleus. The alpha particle, consisting of two protons and two neutrons, is initially at rest and is released from a thorium nucleus with a diameter of 15 femtometers. The relevant equations involve kinetic energy and electric potential, with constants such as Coulomb's constant and the charge of an electron being utilized. The final calculation yields a speed of approximately 4.1 x 10^7 m/s for the alpha particle. The participants emphasize the importance of correctly interpreting the diameter as the radius in their calculations.
Shakenbake158
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Hey guys, I am having a hard time solving for this problem, and I can't seem to figure out what I am doing wrong.

Homework Statement



In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass m=4u and charge q=2e. Suppose a uranium nucleus with 92 protons decays into thorium, with 90 protons, and an alpha particle. The alpha particle is initially at rest at the surface of the thorium nucleus, which is 15 fm in diameter.

What is the speed of the alpha particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest.

Homework Equations



K_{}f+qV_{}f = K_{}i+qV_{}i

V = KQ/R

e=1.6*10^-19

The Attempt at a Solution



K_{}f + 0 = 0 + qV_{}i

1/2MV^2 = q(KQ/R)

R = 1.5*10^-14 meters

K = 8.99*10^9

Q = 90(1.6*10^-19)Coulombs

q = 2e

m = 4u

(1/2)(4u)V^2 = 8630400 m/s

(2u)V^2 = 8630400 m/s

V = 2077.31 m/s
 
Last edited:
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Always show units in you calculations.
 
DONT WORRY I GOT ALL THE ANSWERS! :D:D:D:D:D:D
remember that 15 fm is the diameter not the radius
then just do it :D

Kf = qVi
.5(4u)v^2 = (2e)(k x 90e)/(7.5 fm)
k = 8.99 x 10^9
e = 1.6 x 10^-19
4u = 6.6422 x 10^-27 kg
7.5 fm = .75 x 10^-14 m

solve for v
ANSWER: 4.1 x 10^7 m/s
 
this is a little old :DD
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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