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An electron and a proton are allowed to fall through the separation the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure . (a) Calculate the time of flight for both electron and proton (b) Suppose if a neutron is allowed to fall what is the time of flight ? (c ) Among the three which one will reach the bottom first? ( Take m_(p) = 1.6xx10^(-27)kg , m_(c) = 9.1 xx10^(-31) " kg and g " = 10 ms^(-2)) |
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Answer» Solution :Potential difference between the parallel plates V = 5 V Separation distance h= 1 mm `= 1xx10^(-3)` m MASS of proton `m_(p)= 1.6xx10^(-27) `kg Mass of proton `m_(c) = 9.1 xx10^(-31) `kg Charge of an a proton ( or ) electron E = `1.6xx10^(-19)` C From equation of motion S = `vt+ (1)/(2) at^(2)"" [ u =0 ,s =h]` From equation of motion `h = (1)/(2) at^(2)` `t = sqrt((2h)/(a))` Acceleration of an electron due to electric field ` a= (F)/(m ) = (eF)/(m)"" [ E= (V)/(d)]` (a) Time of flight for both electron and proton `t_(e)=sqrt((2hm_(e))/(eE))=sqrt((2xx1xx10^(-13)xx9.1xx10^(-31)xx10^(-3))/(1.6xx10^(-19)xx5))` `=sqrt((18.2xx10^(-34)xx10^(-3))/(8xx10^(-19)))=sqrt(2.275xx10^(-15)xx10^(-3))` `t_(e)= 1.5 xx10^(-9)s` `t_(e)= 1.5xxns ` `t_(p)= sqrt((2hm_(e))/(eE))=sqrt((2hm_(p).d)/(eV))=sqrt((2xx1xx10^(-3)xx1.6xx10^(-27)xx10^(-3))/(1.6xx10^(-19)xx5))` `=sqrt((2xx10^(-33))/(5xx10^(-19)))=sqrt(0.4xx10^(-14))= 6x32xx10^(-8)` `t_(p)=63xx10^(-9)` `t_(p)= 63 ns ` (b) time of flight of neutron `t_(N)= sqrt((2h)/(g)) = sqrt((2xx1xx10)/(10))= sqrt(0.2xx10^(-3))` `t_(n)= 0.0141 s = 14.1 xx10^(-3)s ` `t_(n) = 14.1 xx10^(-3) ms ` (c) Comparision of VALUES 1,2 and 3 The electron will reach the bottom first |
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