Help With Finding The Speed of an Alpha Particle

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Homework Help Overview

The discussion revolves around calculating the speed of an alpha particle emitted during the alpha decay of a uranium nucleus. The problem involves concepts from nuclear physics, specifically the properties of alpha particles and the energy conservation principles in radioactive decay.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles to find the speed of the alpha particle, while others suggest clarifying the use of units and the interpretation of the diameter of the thorium nucleus.

Discussion Status

Participants are exploring different approaches to the problem, with some providing calculations and others questioning the assumptions made regarding the radius and the application of formulas. There is a mix of attempts to clarify the setup and calculations without reaching a consensus on the final answer.

Contextual Notes

There is a mention of the diameter of the thorium nucleus being 15 fm, which some participants emphasize should be interpreted correctly in calculations. Additionally, the original poster's calculations are noted, but there is no resolution on the accuracy of the final speed derived.

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
 

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