
#1
Mar2908, 05:46 PM

PF Gold
P: 820

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_1q_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_1q_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] 



#2
Mar2908, 06:10 PM

Sci Advisor
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P: 2,886

[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] 



#3
Mar2908, 06:30 PM

PF Gold
P: 820

[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" 



#4
Mar2908, 06:41 PM

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P: 2,886

Particles and charge[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] 



#5
Mar2908, 07:00 PM

PF Gold
P: 820

[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] 



#6
Mar2908, 07:03 PM

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P: 2,886

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]) 



#7
Mar2908, 07:13 PM

PF Gold
P: 820

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] [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 



#8
Mar2908, 08:59 PM

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P: 2,886

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] 



#9
Mar2908, 09:24 PM

PF Gold
P: 820

[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 



#10
Mar2908, 10:15 PM

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P: 2,886

[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 



#11
Mar2908, 10:37 PM

PF Gold
P: 820

[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? 


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