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))`

Explanation

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