Question

An electron (mass m1 = 9.11 x 10-31 kg) and a proton (mass m2 = 1.67 x10-27 kg) attract each other via an electrical force. Suppose that an electron and a proton are released from rest with an initial separation d = 3.50 x 10-6 m. When their separation has decreased to 1.40 x 10-6 m, what is the ratio of (a) the electron's linear momentum magnitude to the proton's linear momentum magnitude, (b) the electron's speed to the proton's speed, and (c) the electron's kinetic energy to the proton's kinetic energy?

Answer #1

Solution: mass of electron = m1 = 9.11*10^{-31}kg.

mass of proton = m2 =1.67*10^{-27}kg.

initial seperation = d = 3.50*10^{-06}m.

final seperation = d1 = 1.40*10^{-06}m.

Now, initial potential potential energy = kq1*q2/d =
9*10^{9}*1.6*10^{-19}*1.6*10^{-19}/3.5*10^{-06.}

^{ }= 6.58286*10^{-23}J.

Final potential energy = kq1*q2/d1 =
9*10^{9}*1.6*10^{-19}*1.6*10^{-19}/1.4*10^{-6.}

^{ }= 1.6457143*10^{-22}J.

So, kinetic energy = final potential energy - initial potential energy.

= 9.874283*10^{-23}J.

m_{e}(v_{e})^{2}/2=
9.874283*10^{-23.}

^{ }v_{e} =
14723.4164m/s.

v_{p} = 343.88208m/s.

(a) Ratio of momentum = m1*v_{e}/m2*v_{p} =
9.11*10^{-31}*14723.4164/1.67*10^{-27}*343.88208 =
0.0233356.

(b) Ratio of speed = v_{e}/v_{p} =
14723.4164/343.88208 = 42.8153.

(c) both have same kinetic energy so ratio of electron kinetic energy to proton kinetic energy is 1.

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