Register to reply

Particles and charge

by ~christina~
Tags: charge, particles
Share this thread:
~christina~
#1
Mar29-08, 05:46 PM
PF Gold
~christina~'s Avatar
P: 824
1. The problem statement, all variables and given/known data
Modern homes that have been tightly sealed for fuel efficiency can have a build up of radon gas inside. This gas diffuses out of the ground and through the foundations of these homes, forming an air pollutant. Radon can decay by emitting an alpha particle (charge ) [tex]Q1= 3.2x10^{-19} C [/tex] In addition to the alpha particle, this transmutation also produces a polonium nucleus (charge[tex] Q_2= 1.3 x10^{-17} C [/tex])

a) find the force exerted on the alpha particle by the polonium nucleus if they are a distance of [tex]d_1= 9.1 x10^{-15}m[/tex] apart. compare your results to the gravitational force that the polonium nucleus exerts on the alpha particle( the masses of the alpha particle and the polonium nucleus are [tex]6.60 x10^ {-27}kg[/tex] and [tex]3.45x10^{-25} kg [/tex] respectively)

b) assume that the polonium is fixed and the alpha particle is free to move. How many MeV's of work is done on the alpha particke as it moves to anew position [tex]d_2= 2di[/tex]

c) If the alpha particle is initially at rest, find it's speed at it's new position

2. Relevant equations
3. The attempt at a solution

a) find the force exerted on the alpha particle by the polonium nucleus if they are a distance of [tex]d_1= 9.1 x10^{-15}m[/tex] apart. compare your results to the gravitational force that the polonium nucleus exerts on the alpha particle( the masses of the alpha particle and the polonium nucleus are [tex]6.60 x10^ {-27}kg[/tex] and [tex]3.45x10^{-25} kg [/tex] respectively)

well since I have

[tex]d_1= 9.1 x10^{-15}m[/tex]

[tex]M_\alpha = 6.60 x10^ {-27}kg[/tex]

[tex]M_p= 3.45x10^{-25} kg [/tex]

I think to find the force exerted on the alpha particle (not sure if this is correct but I couldn't think of anything else)

[tex]F_e= k_e \frac{|q_1||q_2|} {r^2} [/tex]

so is the force just F? I would think so..

so it would be

[tex]F_e= k_e \frac{|q_1||q_2|} {r^2} [/tex]

[tex]k_e= 8.9876x10^9 N*m^2/C^2[/tex]
[tex]Q1= 3.2x10^{-19} C [/tex]
[tex] Q_2= 1.3 x10^{-17} C [/tex]

[tex]F_e= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {9.1 x10^{-15}m^2} =[/tex]
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
nrqed
#2
Mar29-08, 06:10 PM
Sci Advisor
HW Helper
P: 2,954
Quote Quote by ~christina~ View Post
1. The problem statement, all variables and given/known dataI think to find the force exerted on the alpha particle (not sure if this is correct but I couldn't think of anything else)

[tex]F_e= k_e \frac{|q_1||q_2|} {r^2} [/tex]

so is the force just F? I would think so..

so it would be

[tex]F_e= k_e \frac{|q_1||q_2|} {r^2} [/tex]

[tex]k_e= 8.9876x10^9 N*m^2/C^2[/tex]
[tex]Q1= 3.2x10^{-19} C [/tex]
[tex] Q_2= 1.3 x10^{-17} C [/tex]

[tex]F_e= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {9.1 x10^{-15}m^2} =[/tex]
Yes, that will be the electric force except that the entire denominator must be squared (the way you wrote it it looks like you only squared the units) so it should be


[tex]F_e= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)^2} =[/tex][/QUOTE]
~christina~
#3
Mar29-08, 06:30 PM
PF Gold
~christina~'s Avatar
P: 824
Quote Quote by nrqed View Post
Yes, that will be the electric force except that the entire denominator must be squared (the way you wrote it it looks like you only squared the units) so it should be


[tex]F_e= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)^2} =[/tex]
That is what I meant to say. (I had problems submitting things onto the forum before so it wasn't even showing as a fraction and I couldn't edit either) so it would equal

[tex]F_e= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)^2} = 3.44 x10^-64N[/tex]

and when they say to compare it to the gravitational force would they mean to use this:
[tex]F= G\frac{M_1M_2} {r^2} [/tex] thus:

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{ (6.60 x10^ {-27}kg)
(3.25x10^{-25}kg)} {9.1x10^{-15}}= 1.57x10^{-47} N[/tex]

b) assume that the polonium is fixed and the alpha particle is free to move. How many MeV's of work is done on the alpha particle as it moves to a new position [tex]d_2= 2di[/tex]


how would I do this? and what is "MeV's work"

nrqed
#4
Mar29-08, 06:41 PM
Sci Advisor
HW Helper
P: 2,954
Particles and charge

Quote Quote by ~christina~ View Post
That is what I meant to say. (I had problems submitting things onto the forum before so it wasn't even showing as a fraction and I couldn't edit either) so it would equal

[tex]F_e= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)^2} = 3.44 x10^-64N[/tex]
Ok. Looks good then
and when they say to compare it to the gravitational force would they mean to use this:
[tex]F= G\frac{M_1M_2} {r^2} [/tex] thus:

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{ (6.60 x10^ {-27}kg)
(3.25x10^{-25}kg)} {9.1x10^{-15}}^2 [/tex]
Yes. You put the exponent of 2 in the deominator at the wrong place in your tex code but I can see that you mean the correct expression which is
[quote]

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{ (6.60 x10^ {-27}kg)
(3.25x10^{-25}kg)} {(9.1x10^{-15})^2}[/tex]
b) assume that the polonium is fixed and the alpha particle is free to move. How many MeV's of work is done on the alpha particle as it moves to a new position [tex]d_2= 2di[/tex]


how would I do this? and what is "MeV's work"
They want the work done by the electric force (which is essentially the chaneg of electric potential energy, so the difference between [tex] \frac{k q_1 q_2 }{r} [/tex] evaluated at the initial an dthe final points. The result is in Joules if you use SI unites but you can convert Joules to MeV.
~christina~
#5
Mar29-08, 07:00 PM
PF Gold
~christina~'s Avatar
P: 824
Quote Quote by nrqed View Post
Yes. You put the exponent of 2 in the deominator at the wrong place in your tex code but I can see that you mean the correct expression which is
yes, I did mean that, but since I copied and pasted here and there I missed that.

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{(6.60 x10^ {-27}kg)
(3.25x10^{-25}kg)} {(9.1x10^{-15})^2} = 1.727x10^-33 N[/tex]

They want the work done by the electric force (which is essentially the chaneg of electric potential energy, so the difference between [tex] \frac{k q_1 q_2 }{r} [/tex] evaluated at the initial an dthe final points. The result is in Joules if you use SI unites but you can convert Joules to MeV.

hm so you mean I take this number below and subtract it from the distance if it was 2d? like this?

[tex]U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)} = 4.1086x10^{-12} [/tex]

[tex]U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(2(9.1 x10^{-15}m))} =2.054x10^{-12} [/tex]
nrqed
#6
Mar29-08, 07:03 PM
Sci Advisor
HW Helper
P: 2,954
Quote Quote by ~christina~ View Post
yes, I did mean that, but since I copied and pasted here and there I missed that.

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{(6.60 x10^ {-27}kg)
(3.25x10^{-25}kg)} {(9.1x10^{-15})^2= 1.727x10^-33 N}[/tex]




hm so you mean I take this number below and subtract it from the distance if it was 2d? like this?

[tex]U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(9.1 x10^{-15}m)} = 4.1086x10^{-12} [/tex]

[tex]U= (8.9876x10^9 N*m^2/C^2) \frac{|3.2x10^{-19} C ||1.3 x10^{-17} C |} {(2(9.1 x10^{-15}m))} =2.054x10^{-12} [/tex]
Yes (but don't put absolute values...there are no absolute values taken when we calculate potential energy. here it does not make any difference but just to let you know).

And notice that the units are Joules. You will have to convert to MeV (recalling that [tex] 1 eV = 1.60 \times 10^{-19} J [/tex])
~christina~
#7
Mar29-08, 07:13 PM
PF Gold
~christina~'s Avatar
P: 824
I just noticed that this looked funny so I did it over:

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{(6.60 x10^ {-27}kg)(3.25x10^{-25}kg)} {(9.1x10^{-15})^2} = 1.727x10^-33 N[/tex]

Quote Quote by nrqed View Post
Yes (but don't put absolute values...there are no absolute values taken when we calculate potential energy. here it does not make any difference but just to let you know).

And notice that the units are Joules. You will have to convert to MeV (recalling that [tex] 1 eV = 1.60 \times 10^{-19} J [/tex])
Oh..okay. when I convert it it would be

[tex]Fe_i= 4.1086 x 10^{-12}= 2.5678x10^7 eV [/tex]

[tex]Fe_f= 2.054 x 10^{-12}= 1.2837x10^7 eV [/tex]

and subtracting would give me..

[tex]2.5678x10^7 eV - 1.2837x10^7 eV = 1.284125x10^7 eV[/tex]


c) If the alpha particle is initially at rest, find it's speed at it's new position

not sure how to do this either unfortunately.

Thanks
nrqed
#8
Mar29-08, 08:59 PM
Sci Advisor
HW Helper
P: 2,954
Quote Quote by ~christina~ View Post
I just noticed that this looked funny so I did it over:

[tex]F= 6.67x10^{-11}N(m/kg)^2 \frac{(6.60 x10^ {-27}kg)(3.25x10^{-25}kg)} {(9.1x10^{-15})^2} = 1.727x10^-33 N[/tex]



Oh..okay. when I convert it it would be

[tex]Fe_i= 4.1086 x 10^{-12}= 2.5678x10^7 eV [/tex]

[tex]Fe_f= 2.054 x 10^{-12}= 1.2837x10^7 eV [/tex]

and subtracting would give me..

[tex]2.5678x10^7 eV - 1.2837x10^7 eV = 1.284125x10^7 eV[/tex]
Don't forget to convert to MeV (1 MeV = 10^6 eV)
c) If the alpha particle is initially at rest, find it's speed at it's new position

not sure how to do this either unfortunately.

Thanks
Conservation of energy

initial Kinetic energy + initial potential energy = final Kinetic energy plus final potential energy

where potential energy here is again [tex] k q_1 q_2 /r [/tex]
~christina~
#9
Mar29-08, 09:24 PM
PF Gold
~christina~'s Avatar
P: 824
Quote Quote by nrqed View Post
Don't forget to convert to MeV (1 MeV = 10^6 eV)
oh, haha I was wondering what the M was. okay.

Conservation of energy

initial Kinetic energy + initial potential energy = final Kinetic energy plus final potential energy

where potential energy here is again [tex] k q_1 q_2 /r [/tex]
hm..so the potential energy doesn't include the masses of the particles, just charges?

[tex]1/2mv^2 + k \frac{q_1q_2} {r} = 1/2mv^2 + k \frac{q_1q_2} {r} [/tex]

I'm not sure what the equation for the kinetic energy for a particle would be.

Is it like I wrote it, or does it include the 2 masses OR is it some other equation like the potential energy?

Thanks
nrqed
#10
Mar29-08, 10:15 PM
Sci Advisor
HW Helper
P: 2,954
Quote Quote by ~christina~ View Post
oh, haha I was wondering what the M was. okay.



hm..so the potential energy doesn't include the masses of the particles, just charges?

[tex]1/2mv^2 + k \frac{q_1q_2} {r} = 1/2mv^2 + k \frac{q_1q_2} {r} [/tex]

I'm not sure what the equation for the kinetic energy for a particle would be.

Is it like I wrote it, or does it include the 2 masses OR is it some other equation like the potential energy?

Thanks
What you wrote is correct (and yes, the kinetic energy is simply 1/2 mv^2) To be more precise, the equation is



[tex]1/2mv_i^2 + k \frac{q_1q_2} {r_i} = 1/2mv_f^2 + k \frac{q_1q_2} {r_f} [/tex]

where "i" stands for initial and "f" for final.

here I used the fact that only one of the two masses moves (this is why they say to assume that the Po nucleus is staying at rest). So the mass that appears there is the mass of the alpha particle
~christina~
#11
Mar29-08, 10:37 PM
PF Gold
~christina~'s Avatar
P: 824
Quote Quote by nrqed View Post
What you wrote is correct (and yes, the kinetic energy is simply 1/2 mv^2) To be more precise, the equation is

[tex]1/2mv_i^2 + k \frac{q_1q_2} {r_i} = 1/2mv_f^2 + k \frac{q_1q_2} {r_f} [/tex]

where "i" stands for initial and "f" for final.

here I used the fact that only one of the two masses moves (this is why they say to assume that the Po nucleus is staying at rest). So the mass that appears there is the mass of the alpha particle
Oh..okay but you would still include the charges of the two particles, right? Is it because their presence affects the other particle?

[tex]1/2mv_i^2 + k \frac{q_1q_2} {r_i} = 1/2mv_f^2 + k \frac{q_1q_2} {r_f} [/tex]

so since initial v is 0 and plugging in the numbers I got before. Assuming initial is d1 given and the d2 is 2d1

[tex]0 + 4.1086x10^{-12}J = 1/2 (6.60 x10^ {-27}kg) (v_f)^2 + 2.054x10^{-12}J [/tex]

[tex]v_f= 4.99x10^3 m/s[/tex] => is it supposed to be that large?


Register to reply

Related Discussions
2 Electric Charge Particles w/ acceleration? Introductory Physics Homework 7
What is charge? Why do fundamental particles possess it? High Energy, Nuclear, Particle Physics 8
When is a particles charge spread out? Quantum Physics 9
Very small particles induce other particles Quantum Physics 2
Global particles, local particles (Colosi, Rovelli) Beyond the Standard Model 0