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A magnetic field set up using Helmholtz coilsis uniform in a small region in a direction perpendicular to both the axis of the coils and the elctrostatic field. If the beam remains undeflected when the eletrostatic field is 9.0 xx 10^(+5) Vm^(-1)make a simple guess as to what the beam contains . Why is the answer not unique ? |
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Answer» Solution :Since the particle remains undeflected `v=(E)/(B) = (9 XX 10^(-5)xx 100)/(75) = 12 xx 10^(15)= 1.2 xx 10^(-6) ms ^(-1)` `KE - 1/2 mv^2= EV "" therefore e/m = (v^2)/(2V)` `therefore e/m = ((1.2 xx 10^(+))^(2))/(2xx 15 xx 10^3) = ((1.2)^2 xx 10^(12))/(30 xx 10^3) = 4.8 xx 10^7 C kg^(-1)` For deuterium `e/m = (1.6 xx 10^(-19))/(2 xx 1.67 xx 10^(-27)) = 4.8 xx 10^7 C kg^(-1)` Hence the particleis deuterium . Othere particles are `He^(++), Li^(+++)` etc. The answer is not unique because only the RATIO of change of mass is determined . |
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