icctro 11 In the Rutherford model of the hydrogen atom E165.11 In the Ruth (prop
ID: 1430640 • Letter: I
Question
icctro 11 In the Rutherford model of the hydrogen atom E165.11 In the Ruth (proposed by Ernst Rutherford in roughly 1910), an electron is imagined to orbit the proton in a circle whose radius is about ro0.053 nm. (a) Use Newton's second law, Coulomb's law, and what you know about acceleration in circular motion to show that the electron's acceleration in a circular orbit of radius r is : ke2/ mr2 and its kinetic energy is K-ke2/2r, where e = charge on a proton Icharge on an electron A and m is the mass of the electron. El (b) Show t tron in such an orbit is E=-k2/2r Because the electron is accelerating, it will radi- ate energy in the form of electromagnetic waves. Assuming that it does so slowly enough so that its orbit remains essentially circular show that the Larmor formula predicts that the rate at which it radiates energy is (c) (E16.24) 3rmcExplanation / Answer
a) kQq/r*r = mv*v/r (Coulomb's law, newton's second law, centripital force)
v = sqroot(kQq/rm)
a = v*v/r = kQq/r*rm
but Q = q = e
so a = ke^2/mr^2
and K = 0.5mv^2 = ke^2/2r
b) P.E = -ke^2/r
T.E. = -ke^2/r + ke^2/2r = -ke^2/2r
c) By Larmor formaula, radiated power , P = 2ke^2a^2/3c^3
P = 2ke^2 k^2 e^4/3c^3 m^2 r^4
P = 2ke^2c (k e^2/mc^2)^2 /3r^4
d) dE/dt = d(-k e^2/2r)/dt = -k e^2 d(1/2r)/dt = k e^2/2r^2 dr/dt
e) -2ke^2 c ( k e^2 / m c^2)^2 / 3 r^4 = k e ^2 / 2 r^2 dr / dt
-2 c ( k e^2 / m c^2)^2 / 3 r^2 = 1 / 2 dr / dt
-4/3 (k e^2/mc^2)^2 cdt = r^2 dr
f) integrating last part
integral(-4/3 (k e^2/mc^2)^2 cdt)[from 0 to T] = integral(r^2 dr)[from r0 to 0]
4 (k e^2/mc^2)^2 cT = r0^3
T = ro^3(mc^2/ke^2)^2/4c
g) ke^2/mc^2 = 9*10^9 * (1.6*10^-19)^2 / 9.1*10^-31 * 9*10^16 = 2.81*10^-15 m
T = (0.053*10^-9)^3(1.26*10^29)/4*3*10^8 = 1.567*10^-11
Hence rutherford's model is not plausible
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