Consider a proton moving directly towards a nucleus. The nucleus has a charge of

Question

Consider a proton moving directly towards a nucleus. The nucleus has a charge of 35 q_e, where q_e is the charge of the proton. If the kinetic energy of the proton is 39 keV (39 Times 103 electron Volts) when it is a distance d_j = 2.5 Times 10^-11m from the nucleus, what is the kinetic energy when the proton is a distance d_f = d_j/3 from the nucleus? Assume that the nucleus does not move during this time. Give your answer in keV to three digits. Do not include units in your answer. In terms of unit conversions, note that the standard units for energy are Joules, and that 1 eV = 1.6 Times 10^-19 J (the change in potential energy a proton or electron when it moves through a potential diference of one Volt). So, 1 keV = 1.6 Times 10^-16 J.

Explanation / Answer

here,

charge of nucleus , Q = 35 qe

kinetic energy of proton , KEi = 39*10^3 eV

di = 2.5 *10^-11 m

df = 0.83 * 10^-11 m

change in potential energy = change in kinetic energy

change in kinetic energy , KE = k*35*qe*qe/di^ - k*35*qe*qe/df^

KE = 9*10^9 * 35 * (1.6 * 10^-19)^2 /( 2.5 *10^-11) - 9*10^9 * 35 * (1.6 * 10^-19)^2 /( 0.833 *10^-11)

KE = - 6.46 * 10^-16 J

KE = 4.0375 keV

final kinetic energy = KEi - KE

final kinetic energy = 39 keV - 4.0375 keV

final kinetic energy = 34.96 KeV

the kinetic energy when th proton is at a distance of df is 34.96 KeV

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