(a) In one of the landmark experiments in physics at the end of the 19th century

Question

(a) In one of the landmark experiments in physics at the end of the 19th century, J.J. Thomson (1856 - 1940) used electric and magnetic fields to measure the ratio of charge to mass for the electron. In particular, he measured: e/m=1.758820150(44) × 1011C/kg. Starting with the experimental setup, deduce the physical relationship that led to this measurement.

(b) Fifteen years after Thomson’s experiment, the American physicist Robert Millikan succeeded in measuring the charge of the electron precisely. This value, together with the value of e/m, enables us to determine the mass of the electron. The most precise value available at present is: m = 9.10938215(45) × 1031 kg. Starting with the experimental setup, deduce the physical relationship that allowed Millikan to find the charge of the electron.

(c) In the classical model of the atom, we consider electrons to be negatively charged point charges that orbit a positively charged nucleus, similar to planetary orbits. Give a brief qualitative explanation for why this model ultimately fails.

(d) If I bring two positive charges together so that their separation is smaller than the nuclear scale, then release those particles, classical electromagnetism predicts that the particles will fly away from each other faster than the speed of light. Is this prediction valid? Why or why not? Give a brief qualitative reason for your answer.

Explanation / Answer

a) IN JJ Thomsons experiment
he created some electrons from a cathode ray tube asnd accelerated them towards anode which had a phosphoroent material which lightened up where the spot of this electron beam fell.
now in the path of electron, a magnetic field is applied so that the spot moves up or down ( depending upon the polarity of the applied magnertiuc field)
LET THIS MAGNETIC FIELD Be B
now electric field is applied to counter the deviation caused by magnetic field, let this electric field be E
now mass of electron = m, charge = e
so, magnetic force = evB
electrostatic force = eE
equating both from newton's laws
evB = eE + mg
so, let the accelerating voltage of the cathode ray tube be V
then 0.5mv^2 = eV
v = sqroot(2eV/m)

ignoring effect of greavity
vB = E
v = E/B
2eV/m = E^2/B^2
e/m = E^2/2V*B^2 [ E, and B are balancing magnetic and electric fields, V is the potential difference through which the electron is accelerated before it enters the magnetic field region]

b) in millikans oil drop experiment
atomised, charged oil drops are allowed to settle in a chamber at terminal velcity in absence of electric field
so force of drag, Fd = 6*pi*r*neta*v1 (r is radius of the droplet, neta is viscosity of air, v1 is the terminal velocity of oil droplet)
now weight of the oil drop
w = 4*pir^3(rho - rho')g/3 [ volume of oil droplet multiplied by g and the differences of the densities of oil and air]
at terminal velocity, the oil is not accelerating
so, F = w
r^2 = 9*neta*v1/(2g(rho - rho'))
once we know r
the field is turned back on and
qE = 4*pi*r^3(rho - rho')g/3
q = 4*pi*r^3(rho - rho')g/3E

c) in classical model of atom we consider the electrons to be -vely charged point masses orbiting a positively charged nucleus, but this well defined circular motion of a chargerd particle means that this particle should radiate energy as it moves, but this is not confirmed experimentally. hence the moderl of atom (the classical one)_ failed

d) when the positive chqarges are brought closer to each other than the nuclear scale the weak nuclear forces startr to act on the particles and hence the behaviour of the particles is no longer just described by electromagnetism, hence the behavious of one such system will not be as predicted by the electrostatic thoery, but would be different because trhe weak and stron forces are now in play

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