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~christina~
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Homework Statement
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
Homework Equations
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]