An alpha particle (the nucleus of a helium atom, with charge +2e and a mass four

Question

An alpha particle (the nucleus of a helium atom, with charge +2e and a mass four times that of a proton) and an antiproton (which has the same mass as a proton but charge -e) are released from rest a great distance apart. They are oppositely charged, so each accelerates toward the other.
Hint: You'll need to use the two conservation laws. And what does "a great distance" suggest about the initial value of r?

A) What is the speed of the alpha particle when the particles are 1.4 nm apart?
B) What is the speed of the antiproton when the particles are 1.4 nm apart?

Explanation / Answer

Given that the charge of alpha particle is q1= 2 x 1.6 x 10-19 C mass is m1=4 x 1.67 x 10-27kg charge of the anti proton is q2= 1.6 x 10-19 C mass is m2= 1.67 x 10-27kg distance between then is r=1.4 x 10-9m the potential energy between them is U=kq1q2/r =(9 x 109)(2 x 1.6 x 10-19 )(1.6 x 10-19 C)/(1.4 x 10-9) =3.2914 x 10-19 J if v1 is the velocity of alpha particle and v2 isthe velocity of antiproton then from law of conservation of momentum m1v1+m2v2=0 v1=-m2v2/m1       =-v2/4 ...............(1) from law of conservation of energy 1/2 x m1 v12 +1/2 x m2 v22 =U 1/2 x 4m2 x v22/16 +1/2 x m2 v22 =U 1/2 x 4m2 x v22/16 +1/2 x m2 v22 =U 1/8 x m2 v22 +1/2 x m2 v22 =U 0.625 m2 v22 =U 0.625 x 1.67 x 10-27 x v22 =3.2914 x 10-19                                     v2=1.775 x 104m/s using equation 1 v1=0.443x 104m/s     

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