Consider the \"collision\" between two helium nuclei (\"alpha particles\"). Thes
ID: 1773856 • Letter: C
Question
Consider the "collision" between two helium nuclei ("alpha particles"). These particles have a mass equal to four protons and a charge equal to two protons.
Two helium nuclei are heading directly towards each other with the same speed, so v1 = - v2 where v1 and v2 are the velocities of the particles. A detector determines that the speed of the particles is vi = 3.46x10^6 m/s when they are a distance di = 4.80x10^-12 m apart.
If df is the closest distance the particles approach each other, what is the ratio of the initial to final distances, di/df?
Give your answer to at least three significant digits.
Remember: Momentum conservation requires that v1 = - v2 during the entire "collision". (Total momentum is equal to zero.)
Explanation / Answer
From the given question,
let mass of proton =m=1.67*10-27 kg
mass of alpha particle=4m
charge of proton=q=1.6*10-19 C
charge of alpha particle
initial speed(vi)= 3.46x106 m/s
initial distance(di) = 4.80x10-12 m
final distance=df
potential energy= k(2q)(2q)/r
kinetic energy=(1/2)(4m)(v2)
at the closest point, the whole of kinetic energy and potential energy becomes electrostatic potential energy
(1/2) (4m)(vi)2 + (1/2) (4m)(vi)2 + k (2q)(2q)/di= k (2q)(2q)/df
(1/2) (4*1.67*10-27)(3.46x106)2 + (1/2) (4*1.67*10-27)(3.46x106)2 + 9*109* (2*1.6*10-19)(2*1.6*10-19)/(4.80x10-12)= 9*109* (2*1.6*10-19)(2*1.6*10-19)/df
solving we get df=4.8*10-12 m
The ratio of the initial to final distances= di/df
=4.8*10-12/1.15*10-14
=417
The ratio of the initial to final distances= di/df= 417