A positron has a mass 9.11 x 10 -31 kg, and charge q p = +e = +1.60 x 10 -19 C.

Question

A positron has a mass 9.11 x 10-31 kg, and charge qp = +e = +1.60 x 10-19 C. It is moving with a speed of 3.00x106 m/s at a distance of 1.00x10-10 m from an particle (q = +2e, m = 6.66 x 10-27 kg). Due to its size particle stays at rest.
(a) Calculate the speed of positron at 2.00 x 10-10 m from particle.
(b) What is the positron’s speed when it is far away from the particle?
(c) Replace positron with one electron and describe the motion assuming the same speed and direction as given.

Please show me how to solve this problem. Thanks

Explanation / Answer

(A) When positron at distance r form alpha then

PE of positron.

PE = kq1q2/r

at r = 1 x 10^-10

total energy = PE + KE

= [(9 x 10^9 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19)/(1 x 10^-10)] + [ 9.11 x 10^-31 x (3 x 10^6)^2 / 2 ]

= 4.608 x 10^-18 + 4.0995 x 10^-18 = 8.7075 x 10^-18 J

when positron is at r = 2 x 10^-10
PEf = [(9 x 10^9 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19)/(2 x 10^-10)]

= 2.304 x 10^-18 J

applying energy conservation,

8.7075 x 10^-18 = 2.304 x 10^-18 + (9.11 x 10^-31 v^2 /2 )

v = 3.75 x 10^6 m/s

(B) at very away , PEf = 0

applying energy conservation,

8.7075 x 10^-18 = 0 + (9.11 x 10^-31 v^2 /2 )

v = 4.372 x 10^6 m/s

(C) in case of electron PE will be -ve.

total energy = - 4.608 x 10^-18 + 4.0995 x 10^-18

       = 5.085 x 10^-19

when electron is at r = 2 x 10^-10
PEf = -[(9 x 10^9 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19)/(2 x 10^-10)]

= -2.304 x 10^-18 J

applying energy conservation,

5.085 x 10^-19 = -2.304 x 10^-18 + (9.11 x 10^-31 v^2 /2 )

v = 2.48 x 10^6 m/s

at very away , PEf = 0

applying energy conservation,

5.085 x 10^-19 = 0 + (9.11 x 10^-31 v^2 /2 )

v = 1.06 x 10^6 m/s

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